Good evening to everybody. I'm Dr. Himanshu, second year post-graduate president in Mahatma Gandhi Medical College and Research Institute Puducherry. The topic of my paper presentation is breast infiltrating or invasive ductal casinoma with lymph node metastasis, a case study. So in India, breast cancer is the most common cancer in females. Last year in 2020, approximately 180,000 breast cancer cases were reported in India that represented approximately 10% of the total. Breast cancer is a heterogeneous neoplasm of cell clones, each of which has its own growth rate and metastatic potential. Now, histological grade is a good predictor of potential cancer disease progression, but as well as axillary lymph node involvement is another significant feature. It is a strong prognostic factor for invasive ductal casinoma. The two hallmarks that determine the progression from breast ductal casinoma in situ to invasive or infiltrating ductal casinoma are the loss of basement membrane and myoepithelial layer, followed by invasion of the tumor cells into the stroma and the surrounding tissues. I came across a case, 54 year old female patient that came with a chief complaint to our hospital with lump in the right breast from past seven months and retraction of nipples since past one month. The patient also presented with breastlessness which aggravated on lying down. The patient had a history of weight loss and she had a positive history of diabetes malitis, hypertension, tuberculosis and epilepsy. On clinical examination, fullness was noted in the right breast, dilated lines were visible, nipple retraction was positive. However, no scar and sinuses were seen whereas the left breast represents normal. On palpation, the right breast was warm, tender and around 8 into 7 centimeter mass was felt, almost covering the entire quadrants, sparing the outer upper quadrant which was 12 to 8 o'clock position. Beauty orange appearance was also noted. So the lump was hard in consistency but there was no discharge from the nipple. Further on clinical examination of the lymph nodes, right side breast was examined first and on the right side breast it revealed the x-ray lymph nodes were examined, they were palpable and a mass measuring 2 into 2 centimeters which was hard in consistency was felt. The right supra-clavicular nodes were also palpable. There were no palpable abnormalities in the left breast. Now coming to the imaging methods, the surgeons ordered ultrasound the first examination so on ultrasound but we saw an ill-defined inhomogeneous predominantly hypoechoic lesion, a hypoechoic large lesion with cleft-like cystic spaces which was seen involving the breast parenchyma from 12 o'clock to 7 o'clock position. The lesion showed internal vascularity on color Doppler study and as well as predominantly solid areas on elastography. The contrast, after that the contrast CT thorax was done that detailed a lobulated heterogeneously enhancing lesion with some necrotic areas within in the right breast retro-areola region extending to the upper and the lower quadrants measuring 6 into 4.5 into 5.5 centimeter. Along with that some multiple satellite lesions were also noted. The largest was measuring, the largest was seen in the lower quadrant along with that numerous enlarged nodes were also noted in bilateral axillary subclavian supra-clavicular as well as right posterior cervical among which the largest was found in the right axilla. So the CT scan report gave as a carcinoma right breast with multiple satellite lesions with nodal metastasis. So this is the CT scans contrast enhanced thorax image that shows a lobulated heterogeneously enhancing mass with necrotic areas within in the right breast involving the retro-areola region. Ultrason guided FNAC of bilateral axillary lymph nodes and right supra-clavicular lymph node was taken and fine needle aspiration revealed features which were consistent with metastatic lymph node. Further to that of the mass which was revealed ultrasound guided core needle biopsy was done which confirmed the features which were consistent with infiltrating or invasive ductile carcinoma. Further to that the patient was found to be ER positive, PR positive and HER2NU negative. So the patient has an, the patient was found to have an invasive or infiltrating ductile carcinoma that is ER positive, ER positive and HER2NU negative. Coming to the discussion, myoepithelial cells are absent in breast invasive ductile carcinoma. Now P63 marker in immunohistochemistry is used to accurately diagnose invasive breast carcinoma. So the P63 positive myoepithelial cells have been shown to surround healthy epithelial lesions in the breast and the carcinomas in C2 have a ring around the epithelial cells. Infiltrative carcinomas have not shown any staining. In this case we did not observe that the axillary lymph node metastatic breast carcinomas examined had any accompanying myoepithelial cells. And coming to the conclusion, the lack of the myoepithelial coating in metastatic infiltrative ductile carcinoma in regional lymph node tends to be close to that of the primary tumor. The tumor grade can however advance to a higher grade or be downgrade to a lower tumor grade in axillary lymph node. Biomarkers, the immunohistochemistry like ERPR and HER2NEW provides a valuable prognostic information for best therapeutic decisions. These are my references. Thank you so much.


Hello and welcome to the Dave Redan show presented by Buffalo Wobblings. I'm Greg Royce joined by Stevenson track-and-field head coach Dave Redan Coach we're getting right down towards the end of the season First let's talk about some of the kids So let's do now things that you've had they've had some good performances And I put themselves in a strong position to compete at Nationals here in a few weeks. Yeah So right now we're looking at Derek Davis in the hundred right now. He's looking pretty good to get in and then Another male Patrick Watson has a kind of a top time in the 5k and the and the 3k steeple chase Yeah, and I mean some people from last year like Idris and Richie just have I mean I feel like they'll get there It's just I mean we're getting down to the time of year when when they'll compete their best Right and then on the women's side right now Rachel Panix looking really good in the heptathlon. Yeah, and then I Mean we could get a few others to potentially a couple in the shot put with Jess and Danica and then And then maybe Hopefully Kim getting back in the long jump for outdoors as well. Yeah, absolutely. Yeah You just kind of had you know It always seems like there's always like one or two kids that maybe that last weekend put up a big number late Yeah, I qualify to kind of sneak in there So you're hoping obviously as we go through the conference championship in those last chance meets that happens again Yeah, I think that this weekend is gonna be a great weekend I think a lot of people are ready to compete there at their best. Yeah, absolutely So this weekend Matt conference championship I'm you'll be looking to defend, you know, not only your indoor title from the winner You can also your title, you know co-champions winer last year It meets it wider once again and just talk about how that's shaping out for both the man and the woman Yeah, so I mean on the men's side and like you said I mean we're defending champions So we we want to try to do that again But I mean like I say every season it's just kind of like they know that it doesn't just happen They have to to perform well on on this weekend, but I would say it is looking really good I mean if you look at it from a performance standpoint and I mean they're excited So it's gonna be a fun weekend Now the women's side, you know, obviously that the group of throws that you have got you, you know a ton of points in the indoor meet, you know I'm sure you're looking and they've been strong again outdoors So I'm sure you're looking for that as yet as well Yeah, absolutely on the women's side our throwing group has been definitely the The group that scores the most points at the championship meet so we're counting on them again And then the woman I mean it looks like that top three to four teams are pretty close So so just to kind of keep keep improving and have everyone have their best performances this weekend Yeah, I know indoors, you know that second through four group was within, you know, 10 or 15 points So, you know, basically, you know You finish fourth in the winter, you know, hopefully, you know a few more points here there You know, we're up to second or you know, maybe even once for higher But you said it's so close to the top that anything that happened depending on the week And then, you know, obviously on both sides I think what helps you is your depth and just having so many kids qualify for finals and things like that Yeah, and on both sides It's nice to have so many people at the at the top of each event group And that's kind of been a goal of ours is try to keep keep filling in all the different events Yeah, absolutely. All right, coach, we'll thank you once again for your time best select this weekend And we'll chat probably if not before national certainly afterwards. Thanks for all right today we're Dan I'm Greg Royce. This has been the day we're Dan show


आपको ठा आप़ लग है, है नहीं सकते है, है अए वरते बनुआ स्ता तुन से, अव या, अव या या से खारटो हो तू, भोंदी ना आप अगा है, आप पक तुए की बारे है, बहुर लेंगे था, अव या, आप छब है, आसे कसो अथाली, आप घैंचर मेंग लोग लोग � सेपरेशिन is less. I am not talking about the charge separation, the distance between the charge. If you see the number of positive negative charges will be same for all the molecules. So charge separation in terms of distance it is difference is there but number of charges are same. So with charge separation we cannot like the number of charges which is there in this molecule we cannot decide but here you see the distance between positive and negative charges more so we have opposite correct correct one thing you see first of all the first one is most stable right but you see the charge separation is more here but the charges are opposite correct the charges are opposite charge are opposite the opposite charge close to each other are more stable right so third one will be more like charges away from each other are more stable that was the logic right so first and then third after this we have to compare second and fourth right second and fourth you see there are two difference we have here one is number of covalent bond it is more over here and the charge factor right charge is what the negative charge is present on carbon and here the negative charge is present on nitrogen atom right so negative charge if you compare what the first logic that we have which is the number of covalent bond number of covalent bond is more over here right hence the fourth one is more stable than the second one if number of covalent bond is same then we see the charge present on the atom correct negative charge sorry negative charge we will see when we have equal number of covalent bond like you see in the first three cases one three and four the number of covalent bonds are equal right so why we say the first one is most stable because the negative charge is present on most electronegative element that is nitrogen the first one is most stable third and fourth if we compare we have charge separation and opposite charges are there opposite charges are close to each other more stable so third is better than fourth second one will be least because the number of covalent bond is least over here so i think a bit of pain we should have over here fourth one and then the second one is it clear question number three question number three you tell me why are you looking at me because it happens sometimes that man doesn't stop but it doesn't happen to us my girlfriend is not like that wow you are amazing tell me one thing you really killed a man who was killing him निसका जेवाग भी दमो किस देना ही मिल जाएं देवाख से बेख़ा सूँजा तो मैं अपनी अपनी अंबनो से गुड़ाप्टो वला सिब नेखाल रवाग तो काई कि जक तक तो अपनी अंबनो से ब्वला सिप नेखालगे था तो बूस्टी के आगे नहीं बड़़ा। जाई जाई रहा हूँ, बहाँ रहा हूँ जाई दिन तक जी नी बाओगे में। समझे। आगे बड़़ा बड़े है बहाँ लेक लोगा जाए। बैक्रों लोगी लोगी अगी अग या लेक जाई जाए। 4665 minus 3 1143 so that will be minus 5 so this is the hydration energy we have right so if you see this magnitude wise the hydration energy is more right and the molecule first of all when ALCl3 if you put this into water that is in aqua solution it forms what ALCl3 aquas so ALCl3 aquas is what we have AL3 plus ion surrounded by H2O 3cl-cl-ion surrounded by H2O right so when these two ion will remain in ionic ionization in ionic form right then the hydration energy magnitude wise it is more than to that of ionization energy so first of all this molecule will get ionized okay and then the hydration takes place that's the logic right so if the del H of hydration which is del hydration of H if you add this value here ionization energy if this value is negative or we can say if this value is less than zero then the molecule will be in ionic form right which is clear over here minus 5808 plus 5137 this is less than zero hence the molecule will be in ionic form right first the energy required to convert this into ionic form is lesser than the energy released in hydration okay so few like the whole molecule will not convert into aquas form it will remain into the ionic form hence the answer will be option B if this is greater than zero then it will remain covalent next question question number four question number four usually this type of question they do not ask in in any of the exam but you need to know this is there's a just one formula we need to apply and we'll get the percentage ionic corrective okay so that's why I have given you this question that formula if you know you can do this question try this one if you know the formula okay you see the formula here the percentage ionic corrective that we calculate is equals to 16 into here the element is cesium and chlorine right so x cl minus x ces bracket close plus 3.5 into x cl minus x ces square where x cl and x ces are the electronegativity of chlorine and electronegativity of cesium right so you always subtract the less electronegative element here chlorine minus cesium chlorine minus cesium square right now the value is substitute and you will get the answer 16 into 3 minus 0.8 plus 3.5 into 3 minus 0.8 square when you solve this the answer you will get is around 52.14 percent this is the percent is ionic corrected option D is correct so the compound is ionic little bit of ionic character will have this formula yes you can use this formula if electronegativity is given because this electronegativity you cannot memorize for all the elements right that's why these kind of question they do not ask but it is given there so I have given you this question so that you can remember this formula right you must remember this formula nothing is required after this all these value will be given in the question just you have to substitute the value and you'll get the answer understood question number five tell me the answer fifth one is C C is correct Mx3 Sv2 hybridized okay next slide this one sixth one can you see the question or I should mind if I 0 dipole moment D D is not correct check your answer D why not D why D is incorrect and F3 has don't pair overall dipole moment is not 0 SIF has a square planet yeah that's the reason you'll see in all the molecules CLF first of all you see CLF cannot have 0 dipole moment right so we can eliminate this option C is this kind of question you can eliminate option easily CS2Cl2 will not have 0 dipole moment again because any SP3 atom if you have if all atoms or group tasked with SP3 hybridized atom are same then only the dipole moment for that molecule will be 0 SP3 it is but we have chlorine and hydrogen NF3 again will not have 0 dipole moment here because a lone pair present on the nitrogen atom this option we can also remove NF3 if it is clear then this option we can also remove and we are left with only B this is by eliminating option right so you see the common thing over here NF3 if you judge about NF3 whether it is 0 dipole moment or not then either you can neglect and these two or you can select these two like that BF3 has triangle planet geometry right electron deficient molecule no lone pair on it so for this molecule the mu will be 0 dipole moment will be 0 right SIF4 again it is tetrahedral it is SP3 hybridized or what it is DSP2 hybridized you can check that but in all the cases whether it is tetrahedral or square planner right the dipole moment of SIF4 will be 0 hence answer will be option B yeah it's correct so option B is correct question number seven yes yes question number seven which bond angle theta would result the maximum dipole moment of a triatomic molecule xy2 it's simple right when theta is equals to 90 then the resultant dipole moment will be maximum mu1 plus mu2 so it is option A what is the resultant dipole moment formula mu r is equals to mu1 square plus mu2 square plus 2 mu1 mu2 cos theta and we know this thing will be maximum when this cos theta will be maximum and cos theta maximum value is what one which is at mu r will be maximum mu1 square plus mu2 square 2 mu1 mu2 cos theta yeah so okay fine so this is we have so when cos theta is equals to 1 when cos theta is equals to 1 that what happens the value will be mu1 plus mu2 cos theta is equals to 0 which is when theta is equals to 90 so in this case what happens the angle increases from 90 to 180 when the angle increases the value of cos theta will become negative more and more negative right hence the resultant decreases so actually when you have see actually when you have two dipole moment or two vector given like this then the maximum resultant of that vector will be at angle theta is equals to 0 right at theta is equals to 0 the two vector will give you the maximum resultant vector but here we don't have any zero option so we have to choose out of this four right so from theta theta is equals to 90 beyond this value this cos theta becomes negative and here we'll have minus two mu1 mu2 something right that's why given option option a is correct okay otherwise in general if in the board also they ask the two result two vector will have will give maximum resultant when theta is equals to zero mu1 plus mu2 will be the maximum value of two in this case we can say dipole moment or any vector anyways question number eight suppose the observed value of dipole moment of h2o molecule is this what will be the bond angle in h2o dipole moment is given bond angle you have to find out unit of calculation we have in these kind of questions i've given you these questions because usually we don't solve this kind of questions dipole moment related question resultant and all that's why i've given you for practice which will be given if this kind of question they ask cos theta if it is minus minus zero point 2476 theta 104 degree 20 minute this value you should have in this question and you have to do the exact calculation because you see in this kind of question if you get this option then you have to do the exact calculator because you said difference is very small 105 105 20 minute and similarly for this one also 81 is a you are getting all of you are getting a just you have to solve this mu of h2o mu of h2o resultant will be what mu of OH square plus mu of OH square plus 2 mu into mu this mu we have OH OH 2 mu OH and OH cos theta root over of it so this cos theta you have to find out from this because mu OH is given when you solve this cos theta value you'll get this and then the answer will be option a understood question number nine you see question number nine okay what is the answer 9 12 is b b is correct so you have to basically find out the charge on h plus and h plus in both molecule it is given so we know the dipole moment mu this formula again this kind of question we usually do not solve okay dipole moment mu is equals to q into d this d is given for both molecule 127 and 15161 q we have to find out okay the ratio of partial positive charge on hydrogen atom in hcl now right so this d is given so q is equals to what mu by d and mu is given 1.03 d by it is right so when you multiply this with 10 to the power minus 18 it converts into cgs system the unit is electrostatic unit centimeter this divided by the bond length it is 127 picometer and when you multiply it by 10 to the power minus 10 it also becomes centimeter right so when you solve this you'll get 0.81 into 10 to the power minus 10 esu cgs system it is what we can say 1 d by unit is equals to 10 to the power minus 18 insidious system okay so q by e if you calculate the fractional charge i'm solving only for this next one also you can do it in the same way the fractional charge you have to calculate so that will be equals to q by e electrons and insidious system the charge on electron we can write it as 0.81 10 to the power minus 10 as it is divided by 4.8 into 10 to the power minus 10 insidious system so you solve this and you'll get 0.168 approximately similarly you do the same way and calculate fractional charge for h i take the ratio the answer you will get is around 3.42 is to be option that is what you are getting all of you so this is how we solve this kind of question not that tough or there is no much concept into this but you must solve one or two questions like this so that if it comes you can do it easily right next question number solve this covalent character we'll see what we'll see simply the charge right phasans rule so when charge is mode covalent character is mode so here we have plus here we have the charge in the kcl we have one csl to we have two three and four so according to the phasans rule s i cl four will have the maximum covalent characteristics and kcl will have the minimum so you see the option sorry see option b is correct over here all of you have done okay question number 11 you solve solve question number 11 born energy is given resonance energy you have to find out this question they ask to calculate the resonance energy tell me what happened this week is getting b you see the formula of resonance energy just a second just a second resonance energy you need to use this formula only one formula we have that is l of suppose the molecule is hf here so del of hf will be equal to the bond energy of the molecule given here in this case it is hf plus sorry minus root over of the bond energy of h2 from which the molecule has been formed and the bond energy of f2 this into this okay so the bond energy of hf is what it is given everything you know it is given in the question it is 135 minus bond energy of h2 38 bond energy of f2 135 this kind of question you will have sorry it is h2 it is 104 right for h2 it is 104 and for f2 it is 838 so when you solve this you will get 72.14 whatever the unit we have that will be as it is answer will be option c right answer will be option c so these kind of question like this like when you have few days left seven eight ten days if you have left just you try to see these kind of question also which does not require does not need any logic they just need you to you just want you to know this particular formula all the data will be given in the question just use those data use this formula you will get the answer okay you try to collect these kind of any chapter any subject if you are doing try to collect this kind of information in the last few days for benzene again you have to calculate the number of see for benzene it is different for that you need you need two data here this is the benzene so for that you required carbon carbon single bond and carbon carbon double bond this bond energy it requires and i have i have done this in thermal dynamics i have given you this thing how to find out the resonance energy over there so for that you will understand because it is the average value of this so we'll take the average of all these here carbon hydrogen bond carbon carbon bond carbon double bond carbon okay but in organic chemistry they won't ask you to find out the resonance energy right they will ask some concept related some resonance concept related question right but this kind of resonance energy they will give you here in this chemical bonding chapter okay when you remember this formula with this given data you can find it out right okay so this is how we do we don't have actually any expression here right but with with when you have the question of benzene they'll give you something called empirical resonance energy and hybrid energy will also give it right so we need to take the difference of those two you will understand that only difference you have to take but for that you required some data we don't have such formula for benzene they'll give you empirical resonance energy they'll give you hybrid energy they'll give you carbon carbon single bond double bond with that we can calculate either they'll give you this or they'll give you simply hydration and simply the energy of the hybrid structure and the empirical resonance energy from that difference you can find out but usually with resonance since we don't have such formula such expression they don't ask a resonance energy to calculate for benzene okay next question you solve question number 12 its smallest bond angle you have to find it or i think it's clear the smallest bond angle ns3 ns3 tell me what is the reason tell me the reason why ns3 will have the smallest bond angle okay one lone pair what about h2o and h2s how many lone pair see first of all when you see the fourth option is so2 which has linear geometry right so this cannot have the minimum bond angle and all these three options has is sp3 hybridized so ideally if there is no lone pair the bond angle should be 198 109 28 degree correct ideally but since lone pair we have present then the bond angle will be less than this because of the lone pair bond pair repulsion as the number of lone pair increases bond angle also decreases because if you have this lone pair present here right this will suppress these bonds right so hence the bond angle will decrease so obviously since h2o and h2s will have two lone pair of electron the bond angle in these two will be lesser than to that of ns3 is it clear right so we are left with option a and b trying a and b if you see we have h2o and h2s first of all the electronegativity of oxygen is more than to that of sulfur right and the size of sulfur is more than to that of oxygen right so because of size factor and less electronegativity also here because the electronegativity of sulfur is less the two bond two lone pair of electron will suppress more to this bond here we have but since the electronegativity of oxygen is more here so this lone pair is very close to the oxygen atom and hence the effect of this lone pair to suppress this angle will be lesser in comparison right that's why option b is correct over here did you understand this this lone pair will suppress this bond pair of electron if the electronegativity is less if the electronegativity of central atom is more this lone pair is more close to the oxygen atom and hence the effect to suppress this bond pair will be lesser see what i said suppose we have any compound like this and here we have two lone pair present so this lone pair will suppress this bond pair of electron and hence the bond angle will reduce right but the effect of this lone pair on this bond pair of electron will be less if the electronegativity of central atom increases why because if the electronegativity is more this lone pair of electron is very close to this central atom right and hence the effect of this to suppress this bond pair of electron will be less comparatively less right so what we can say as the lone as the excuse me electronegativity of central atom decreases excuse me electronegativity of central atom decreases the lone pair will suppress the bond pair of electron more right and hence the bond angle decreases that's why it is minimum for option b is it clear you should not choose this c option that that was a you know shock for me when you said when all and when all of you said option c is correct you can easily see hybridization is sp3 one lone pair two lone pair two lone pair so lone pair bond pair repulsion will be more over here so less will be the bond angle so any one of these two you can select okay but if you when you said this it was a actually shock for me anyway so you take care of all this thing okay option b is correct over here you can also memorize in another way as the size of the central atom increases bond angle decreases in the same group can we move on ram charan is getting a with the most of you are getting a sanjana spectrum mathly a is correct so i'm not discussing this okay you can easily find out the static number calculate the electron divided by 8 and then you can find the static number the answer is correct x e f 2 only one i'll calculate 8 plus 7 into 2 divided by 8 so it is it has two bond pair and three lone pair right so 3 should be the first one so d option is correct so only you can calculate for others right next one 22 all of you are getting b option b is correct do i need to solve this no i guess all of you have done right 23 anti or anti prismatic see anti prismatic actually it is not in our slivers okay but if when you have since you have asked let you know the square anti prism is actually it is an there are infinite number of triangles is there the base will be in square right but there are infinite number of triangles okay okay which is in set of the anti prism and the number of triangles should be should must be the even number of triangle it is right so even number sequence of triangle sides close by two polygon caps okay it is difficult to understand like this you just google it okay you will see the you know the structure actually of anti prismatic okay then you will understand will actually it is it has a square and then with the square we have number of triangles are there which is attached by the sides okay and the number of triangles must be even number of triangles that's the only thing we have in this okay so it is difficult it is difficult to you know excuse me understand theoretically but once you see the image you will probably understand okay so better you google it you will understand that 23rd most of you are getting a the molecule shape this r right so you see this kind of questions you can solve easily 23rd a is correct i am not discussing this also 24th you tell me 24th polar and sp2 hybridization b and c is obviously not the answer and the structure of hcl o2 is what it is cl double bond o and o h okay but the structure of s2co3 is this c double bond o o h diabasic acid sp2 hybridization and polar also b is this cl double bond o o h understood hcl o2 is an acid right so h is attached with oxygen over there and cl with double bond o option a is correct fine question number 25 the number of bond pair and lone pair i3 minus is important it has been asked in the exam okay some of you are getting a some of you are getting b okay you tell me what is the geometry and shape of i3 minus shape is linear what is geometry shape is linear it's correct geometry you tell me electrons i3 minus has 7 into 3 plus 1 divided by 8 so it has 2 bond pair and 3 lone pair right so obviously 3 and 2 option a is correct right the shape will be linear and geometry will be trigonal bipyramidal it is like this we have one lone pair here another lone pair here and another lone pair here i will have here i and i and this is forming a triangle here like this this is the triangle and this is the three i here three lone pair at the vertex of this triangle right geometry is trigonal bipyramidal shape is linear three lone pair and two bond pair question number 26 hybridization all of you are getting b an option b is correct 26 option b 27 the orbital involved in sp3d hybridization what is the geometry of sp3d dz square orbital is there the geometry must be octahedral or square bipyramidal because the z square orbital has two lobe right like this if it is taking part in hybridization this must be also like involved in bonding so the geometry should be square bipyramidal when dz square is there right this is actually a factual base question the answer for sp3d hybridization is dzx in this orbital dzx d orbital is involved if the d sp2 orbital is there then the d orbital involved is dx square minus y square okay sp3d dzx dz square involved in octahedral complex sp3d 2 answer is option b understood can you move on question number 34 which are the following spc's does not exist we calculate bond order in this right if bond order is zero the molecule does not exist option c is correct this one kinetic molecule super oxide why some of you are saying a option super oxide option c is correct oxygen molecule o2 we have already seen this is paramagnetic molecular orbital theory right paramagnetic right super oxide ion is o2 minus this is also paramagnetic 18 to 2 plus 1 17 electrons this is also paramagnetic carbon molecule is c2 this is diamagnetic you can check and then unipositive nitrogen molecule that is n2 plus this is also also has 13 electron so it is also paramagnetic so option c is correct in this yes 36 36 one point order condonia is getting b i have discussed this yesterday or day before i think sanjana is getting b mackley is getting b yeah yeah exactly not the same question i think the same question about okay so here we have intermolecular hydrogen bonding right intermolecular hydrogen bonding and for intermolecular hydrogen bonding the order will be this maximum we have for para then we have meta and then we have ortho so one four will have maximum para one three will have meta three two one and then four right three two one and four option b is correct yes 37 strongest hydrogen bond in vapor phase is strongest hydrogen bond is possible when we have maximum electronegativity difference right that is possible in in the case of fluorine and hydrogen option is correct okay next question number 38 question number 38 oxygen hydrogen right the most electronegativity difference we have so 38 option c is correct oxygen hydrogen will have the most electronegativity difference t is not possible you see we have hydrogen molecule here put with so we don't have any chart separation here it is covalent completely covalent that's what h2 molecule see when hydrogen bonding is possible when we have ho h and another molecule will have hydrogen if there is a chart separation this should have positive and this should have negative charge then only this sorry so positive we should require some negative charge here but in case of h2 will have h h bond right we don't have any chart separation in these two that's why this hydrogen bonding is not possible correct okay question number 39 39 this question is also based on hydrogen bonding most of you are getting c some of you are getting b d why c is the answer the answer is c it's right but why c see k h f2 molecule yeah c k h f2 molecule will exist like k plus and h f2 minus this negative charge will be there on the one of the fluorine atom here right so this f minus will form hydrogen bonding with this hf right that's why this molecule exist in this form when kf combines with hf to form this and because the hydrogen bonding it exist in hf2 minus and k plus will be there yes correct that that's why hf2 minus will have so this is also one another type of question we have which is based on hydrogen bonding again any bond is stronger yes any bond is stronger here always stronger than the hydrogen bond see what happens here kf will actually exist in k plus and f minus and this f minus because of this electronegative difference here we already have h plus and f minus so this also has tendency to make a hydrogen bond here right and the molecule will exist as k plus and hf2 minus kf is in strong electrolyte right like k cl it will exist in ion forms and the second molecule hf electronegative difference is there to charge separation will be there and hence the hydrogen bonding is there right option c is correct question number 40 what is the answer c is obviously correct chemical bond always forms when the force of attraction is more than the force of repulsion then only it will overcome right absolutely pure water does not contain any ions not true it has ions h plus and oh minus will be there because of the electronegativity difference okay so option c is correct here hf is obviously more polar than hbr in covalency there will be sharing of electron not transfer of electron so c option is correct 40 question number 40 option c question number 41 metallic luster exhibited by sodium is explained by proton is not possible diffusion of ions also not possible metallic less is because of the electrons right only one option is there oscillation of loose means what the movement of the electrons or loose electrons obviously option c is correct here next one these are the previous year questions yeah it causes color but here here the question is metallic luster right so the metallic properties of any metal we have that is because of the electrons not ions okay electrons not ket ions or positive ions okay i'll see the previous question the question is what metallic luster means the metallic one the metallic layer that you have or you can say the metallic properties of that any metal if you have that is because of what reason that's the question color thing is there because of the ions right but metallic nature is only because of electrons because of movement of electron that's why it is oscillation of loose electrons movement of electron only we have metallic nature right okay now you see the next one question number 46 46 46 is b it's correct right because you see this i3- is made up of is made up of one iodine molecule and i- linear just now we have discussed shape is linear and with this i- this cs plus has joined right so that's the complex compound it is kind of the complex compound we have that's why option b is correct you must keep this in mind this is important not three minus you three i minus you right okay 46 one is b what about 47 few questions we have already discussed in this you must try to figure it out like what kind of question they ask in the exam 47 you are getting b if i is unsymmetrical right but it's pentagonal pyramidal it is pentagonal shape we have of this right so 47 option b is correct all of you are getting option b 48 spf what is the shape what is the to try to find out the geometry of spf 5 tell me the geometry of spf 5 all of you what happened spf 5 the diagonal bipyramid yeah it's correct triangle bipyramid and that's why we have symmetry over here right and because of this symmetry the dipole moment net dipole moment will be zero right so symmetry we are just checking 48 48 yes what is the answer question number 48 i f 5 you're talking about what is more than i would invite i'm here it is covalent isn't that how you find body point and don't burn all polar means dipole moment not okay see in this i f 5 see in this i f 5 there will be a bit of see if the molecule is covalent right if the molecule is covalent but still if they have electronegativity difference because you must have seen the ionic nature and covalent nature the electronegativity difference will have of specific value more than that it will be ionic less than that it will be covalent right but in that also we have few exceptions right if the bond is covalent also but still it is iodine and fluorine we have huge electronegativity difference and size difference also right so because of the both factor size and electronegativity is still we have little bit of you know charge separation that will give you the dipole moment which is not net dipole moment is not zero here because the geometry is not symmetrical that's the reason see one bond can be polar or non-polar but when you are talking about a molecule we have to see whether we have because of the one bond will have dipole moment right but for dipole moment for a molecule if you are talking about then we have to check whether we have net dipole moment or not right so in that you have to consider all the bonds which is present in the molecule question number 48 option c is correct it has 4 d pi p pi bond the structure of x e o 4 is this x e double bond o this is a structure so d p 4 d pi p pi bond is there option c is correct what about 49 how do we do this option b is correct so how did you get option b sanjana how did you get option b ram chiran how atmesh kushal all of you are getting option b how is it option b see it says that bond order is same but they are not equally stable so in this type of questions we'll just check how many electrons how many electrons are present in anti bonding molecular orbital so the concept is what as the number of electrons in anti bonding molecular orbital increases stability decreases that is what you have to find out stability decreases yes anti bonding electron so you just write down the electronic configuration of h2 it has 2 electron so sigma 1 is 2 for li2 it has 6 electron so sigma 1 is 2 sigma star 1 is 2 sigma 2 is 2 then we have b2 10 electron so sigma 1 is 2 sigma star obviously in b2 we have more anti bonding electrons right so maximum anti bonding electron so stability of b2 will be minimum and h2 will be maximum so option b is correct number of anti bonding electrons we'll check question number 50 50 and 51 both you tell me 50 is c c is correct if you have any doubt you can ask me okay i'm not doing this kind of question because these are very basic and easy questions we have 50 is c it's correct what about 51 the shape of i f6 minus 50 so 50 by 8 6 and 1 7 so it's not pyramidal it's not octahedral it's not square antiprism it is distorted octahedral triangular distorted octahedral hybridization option a is correct what is the hybridization you tell me you see for square antiprism the steric number should be 8 then only it is square antiprism and if steric number is 8 it also involves f orbital not f it will be sp3 see anything is possible you cannot say sp3 d3 is fine till 7 but after that you have to draw the like you know electronic configuration and then according to the compound you'll see what is the next orbital we have which takes part into the hybridization so usually we we discuss only till sp3 d3 beyond that is not in our syllabus you will not get questions of this but since the question is given it is enough that you should know that if the steric number is 8 then only the shape will be square antiprism and with 8 steric number we have sp3 d3 7 orbitals are there what will be the 8th orbital it depends on the molecule according to its electronic configuration according to that only we'll see which is the next orbital which can take part into hybridization so that its steric number should be 8 so according to that you have to think right so like this we cannot say maybe f orbital is there maybe we have some inner orbital also involve anything if possible so you cannot say like that but if you have 8 steric number square antiprism you can choose which is not obviously in this case because the steric number is 7 you are getting here and with one lone pair right that's why it is diagonally distorted octahedral is correct 52 i have discussed this question 52 i have discussed see all these contains weak filled again dsp2 it is actually a question of what we say coordination compound but there also we have hybridization so this may ask in may given this particular chapter also all these are weak filled again dsp2 is possible when we have pairing against the whole stool and that is possible in the case of strong filled again right but since all these are weak but in case of palladium what happens because the large size of pd2 plus the pairing is possible i have discussed this that day when we were discussing coordination compound so here the answer will be option a yes we have discussed this because the size of the palladium and 51 the hybridization see i have 6 minus the number of electrons here we have you see that is 7 into 7 plus 1 divided by 8 that will be 50 by 8 the 6 bond pair and one lone pair so hybridization will be sp3 t3 6 bond pair and one lone pair sp3 t3 for option a for this the last one we have only the information that its steric number should be 8 hybridization it depends on the compound we have 52 is a we have discussed this right you must remember this particular question large size of palladium leads to the pairing of electron against the whole stool if the ligand is weak ligand also 53rd okay we'll start from this question 53rd you let it be not 53rd we'll start from here after the break we'll take a break i'll take my break first now okay so we'll start in 15 minutes 11 30 we'll start okay fine let's know weekly 11 30 we'll start will happen or more okay we'll start 11 30 okay can we start now 53 is d okay answer is correct d option but here we use different formula because here the bond energy rule is not given right but bond length is given and then dipole moment is given okay so in the form of in with dipole moment the expression of percentage ionic character i'll just write down the formula can be given as mu observed divided by mu calculated into 100 how do we calculate this mu this mu we can calculate mu observed is given in the question it will be e electron into distance electron chance we know already 1.6 into 10 to the power minus 19 and the distance is also given 9.17 into 10 to the power minus 11 okay in all these kind of question you must you must take care of units okay so this thing will use and when you substitute this year mu observer is given will get percentage ionic character option d you will get as the answer okay so this is one formula and another one we have also discussed those these two formula you must keep in mind to calculate percentage ionic character maybe on the previous formula they may ask you some question because this formula they have asked once in 13 so the previous one they may ask you one question on that okay so you must keep that in mind stability of this piece of this how do we do this 54th tell me the answer 54th tell me the answer 54th 54th you are getting a 54th a is correct we have discussed this kind of question what we'll do we'll calculate the bond order correct more bond order mode will be the stability if bond order is equal right then we see the number of electrons in anti-bonding orbital if the number of electrons in anti-bonding orbital is less is more stability will be less do i need to solve this 54th a is correct option a you want me to solve this then let me know please see li2 minus and li2 plus has equal bond order right right so on the area equal bond order so when the bond order is equal just you check the number of electrons present in anti-bonding molecular orbital if the electrons present in anti-bonding molecular molecular orbital is more stability will be less understood so in li2 plus if you see there are two electrons present in anti-bonding molecular orbital but li2 minus there are three electrons so stability of li2 minus is less than to that of li2 plus correct next question number 55 d we have just now we have done what about a purvik how it is b dithvik how it is b oxygen is paramagnetic in nature that is how we get molecular orbital theory purvik and dithvik how it is b oxygen has 16 electrons fine but when you draw the electronic configuration according to the molecular orbital theory oxygen molecule is found to be paramagnetic in nature it has unpaired electron and that is how we have got the molecular orbital theory yes a and d are correct you see the question is what is are expected to be exhibited right question your answer was correct a and d n2 and c2 you draw the molecular orbital no it doesn't mean that is expected to be o2 and s2 both are paramagnetic it simply means that which of these are diamagnetic in nature right so you draw the you draw the electronic configuration of n2 according to mo2 n2 and n2 has 14 electrons you draw the structure you will understand c2 we have just now we have seen just you check for nitrogen o2 we already know it is paramagnetic and so s2 also both belongs to the same group no issue purvik tell me n2 is paramagnetic yeah so a and d are correct n2 and c2 54th one is a 55th one is a and d so whenever you have this kind of question is r when it is there you have to check all the options okay take care of this yeah so n2 and c2 are correct a and d tell me question number 56 again not likely to exist means what bond order should be zero you can calculate the bond order easily option c is correct 57 57 is b 57 also correct so you see in all these question which has been asked in jxm you see none of these are tough all are easy questions and whoever made some mistake here that is the mistake has been made that those are casual mistake actually so when you read out the question properly see the options properly what is the question saying what could be the answer like in this one 55th one is r it is there so you have to check all the options maybe more more than one correct in this kind of question do you have to take care of this kind of questions right so basically the point i'm trying to make is what we have solved around 50 questions in this more than 50 i guess 50 to 60 question we have solved and you must have felt that these 15 questions or 10 12 question that we have done just now previous questions these are the most easiest questions we have done right so it is not like that they ask tough questions in the exam okay just you read out the question properly and you have to be very focused right basic thing you should know and while marking the answer you have to be very focused on what is the question is there more than one correct is there or not what is the exact language of the question so you better you take care of those things don't make any silly mistakes usually because of silly mistakes only you have to suffer in the last okay so when you are solving some questions try to avoid those mistakes calculation mistakes and all like you said got confused with the language of the question so to understand the question also is a kind of test that the examiner take okay so you have to understand the question then you have to solve the question and get the right answer anyways so this is for this chemical bonding we have discussed we'll solve some questions on atomic structure now this one vashnavi is getting d is it d vashnavi check your calculation check your calculation sanjana is getting a vashnavi is again getting b b is the correct answer vashnavi sanjana you check your calculation okay you see the critical wavelength for producing the photoelectric is this critical wavelength means what this is the wavelength corresponding to the work function of the metal which is nothing but the lambda not okay now the equation of photoelectric effect is what h nu incident is equals to h nu not plus the k max this is nothing but the work function so kinetic energy is what nu in we can write h c by suppose lambda one minus h c by lambda not so according to the question this lambda not is given 2600 what wavelength should be necessary to produce photoelectrons from prangasthan having twice the kinetic energy those produce at 2200 angstrom right so the kinetic energy suppose according to the question at wavelength lambda one is equals to the kinetic energy wavelength lambda two where the lambda two is what lambda two is 2200 angstrom according to the question right now this will substitute here so here the kinetic energy will be h c by lambda one minus h c by lambda not sorry you have two into this according to the question so two into h c by lambda two minus h c by lambda not right this is what the equation we have all these h c will get cancelled right lambda two is given 2200 lambda not is given 2600 this is also given 2600 this you have to find out lambda one so this equation you have to solve and when you solve this lambda one you will get 1907 angstrom option b is correct so you see one more thing here in this question that you see the options are very close b and c so you don't take any approximation here right do the exact calculation when the options are close like this okay question number 10 all of you are getting a tell me others option a is correct and if you want me to solve let me know otherwise since most of you are getting i will not solve it but you let me know any one of you if you have any doubt you let me know i will solve it and you require to stop the ejection of electron from cu plate is 0.24 the radiation is this nanometer strikes the plate the work function is stopping potential is given thus you need to use the equation calculate the energy that is incident onto the plate which is nothing but hc by lambda lambda is given that energy is incident that energy is incident from this we'll calculate energy e which is equals to hc by lambda this energy and out of this energy this energy is used for the stopping potential so work function plus this is equals to this energy right so whatever this you are getting this minus this will give you the work function one thing you have to take care here when you use this expression e is equals to hc by lambda right the energy you will get in jul but since this stopping potential is given in electron world so either this you have to convert in jul or this you have to convert in electron world since the option is given in electron world so best way is to convert this into electron world right so one jul for this if you have to convert into electron world e we have to divide it by 1.602 into 10 to the power minus 19 this becomes the electron world now this minus 0.24 electron world gives you the answer so answer you will get is option e 11 also you have done question number 11 25 watt bulb it's a very common question okay you have to find out the rate of emission quantum per second 11th you are getting a option e is correct energy will calculate hc by lambda right and the watt and the power is equals to what the number of photons emitted per second into the energy of the photon right so whatever value we'll calculate that is nothing but the power and that will divide by the energy of the photon you'll give you the quantum per second right option a is correct see in atomic structure unit is very important you know everything but once you miss out the unit thing you'll get a wrong answer and that will be there in the option the unit you must take care of the convergent thing you must see and all those things you have to keep in mind next one so these are the the three questions that we have discussed the three different kind of question but the concept is what concept is photoelectric effect only okay so those three different kind of question they ask in photoelectric effect okay so that is what you have to keep in mind question number 19 19th you are getting b right only formula you have to use option b question number 20 question number 20 question number 20 again you are getting b all of you are getting b so i will not again solve this b is correct question number 21 what is the n value for second excited state n is equals to three okay fine calculate e value is given one mole of hydrogen atom means any atoms we have to shift right so option you see the energy is given in electron volt option is given in kilo joule so that electron volt you convert into kilo joule first joule first rubik is getting d correct put with d option is correct so first of all this 12.084 we have to convert into joule and that will be 1.6 into 10 to the power minus 19 this is joule for one atom per atom one mole of atom we have means any atom means this we have to multiply by n a right so 12.084 into 1.6 into 10 to the power minus 19 into n a avocado number this is joule per mole and then we have to convert this into kilo joule in the last answer you will get 1164 kilo joule only unit conversion we have here like i said unit conversion is very important in this chapter question number 22 question number 22 what is the value of n1 and n2 for lineman series shortest wavelength n2 is what the final is equals to what and what is nf lineman series the electron comes into first energy level right n is equals to 1 so nf is 1 and since shortest wavelength we have to find out and i should be the maximum value which is nothing but infinity so 1 by 1 by lambda minimum is equals to r h z square why z square because we have helium here he plus divided by 1 by 1 square minus 1 by infinity square whatever z value is what for achieve plus z value is 1 so when you substitute this r h you will get from this r h is equals to 1 by x according to the question for the same thing we use for barma series okay it is for hydrogen atom so it's 1 fine he plus so for barma series the wavelength is not given we have to find out so longest wavelength 1 by lambda max is equals to r h z square for barma series it is 1 by 2 square minus 1 by 3 square nf minus ni r h is given 1 by x and when you solve this you will get 1 by lambda max is equals to 5 by 9x so lambda x will be 9x by 5 22 option a yeah option a is correct question number 23 question number 23 what is the velocity and orbit relation v in nth orbit is inversely proportional to n right so what will write v in first orbit divided by v in third orbit it will be 3 by 1 right so what we have to find out this is written on third orbit so v3 is equals to what v1 by 3 so x by 3 it is so option b is correct 24th user for energy formula e2 minus e1 e2 minus e1 you have to find out 24th one see the unit is joule per mole so accordingly you will write the expression for energy 9.8 4905 it's correct 24th b is correct only formula you have to use any one of you if you want me to solve let me know and do this yeah what happened where is ramcharan did you get the answer ramcharan okay you see joule per mole we have here the energy formula we have in joule per mole if i write so en is equals to minus 1.312 into 10 to the power 6 by n square because you see here what is given the ionization energy is this ionization means what one electron you have to take from since we have this ionization energy so we are talking about first ionization energy here so from e1 to e infinity you have to take means from the first orbit electron you have to take at infinity that is nothing but the removal of electron or ionization energy so that will be e infinity will be what 0 minus e1 so that will be minus e1 and which form value is given is minus 1.312 into 10 to the power 6 joule per mole so in en this divided by n square will be the energy now we have to find out e2 minus e1 right the energy required to excite in electron in atom from 1 to 2 so from first orbit to second orbit you have to go so whatever the energy difference we have here that we have to find out so what will write e2 minus e1 is equals to what this this formula will use minus 1.312 into 10 to the power 6 open bracket 1 by 2 square minus 1 by 1 square this is what you have to solve right and answer you will get approximately around option b understood vashnavi last question 25th one last one for today 25th the wavelength of a neutron with a translateric kinetic energy equal to this at 300 kelvin is what is k k is not mentioned over k is bolsman constant kt k is bolsman constant you see in this one k is the bolsman constant it's not mentioned here but it is a bolsman constant so kinetic energy it is given that is kt where t is the temperature this will be equals to bolsman constant is r by n not n a into temperature now r value you know 8.314 temperature is 300 and any value you have 6.02 into 10 to the power 23 and when you solve this you'll get the value of kinetic energy is 4.143 into 10 to the power minus 21 june this is the kinetic energy and we know the mass of a neutron is what it is 1.67 into 10 to the power minus 27 kg unit you must take care of again i am telling this okay and kinetic energy we know but we have to find out we have to find out the wavelength so wavelength is equals to what h by mv h we have right and this we can convert into in the form of kinetic energy and that will be h by 2m into kinetic energy root over of it in this you see everything you have you have kinetic energy you have h value you have mass to put this and you will get the answer the answer will be approximately 178 pico meter understood this so we'll wind up the class here only okay tomorrow if you have class chemistry class i will let you know because i have to take 11th class tomorrow class 11 i have to take tomorrow so if you have class i will let you know otherwise we'll send you the schedule by the day okay whenever you have class we'll send you the schedule right in organic you must solve i have i have sent the questions on d and f block element that you must have seen that there are sections different different sections like this is for you know the compounds of iron this is for the magnetic properties k m m of 4 so on the basis of that they have given few questions section wise okay so i have given that question i have sent that questions with answer key also you must go through with that in organic every day you read three four pages now also with some questions must you see because you never know suppose last day also if you read one particular line they may ask that particular line only in the exam okay so last days five ten days that whatever you have they try to get the gather the information that is it okay don't try to read or understand any new concept okay so fine we'll wind up the class here only see you soon okay bye bye


What's up guys, my name is Michael and welcome to my YouTube channel today we are going to go over a problem called the next palindrome and it's an ad hoc problem. So it's a problem that we're just requires you to you don't need to know anything about algorithms you just need to think and solve the problem. So it's a problem about thinking. Okay, so recall what I said about a palindrome a palindrome is let's say we have a number of two, no, not two, nine, two, nine, okay. And a palindrome is when the left most number is equal to the right most number that's basically what a palindrome is. So like if for all the numbers, the left one is equal to the right one, okay. So if in this case, nine is equal to the right most number, not nine. So it is still considered a palindrome and two is equal to the itself to is equal to itself. So it's also a palindrome. So this is a palindrome. Okay. Now, another number two, one, three, two, this is not a palindrome, although two is equal to the right most value of two, one is not equal to the right most value of three, right? One is not equal to three, right? So the a palindrome is basically if you were to reverse the numbers up to the middle and they're equal, then that means that it's a palindrome. So this is not a palindrome where one is not equal to three, like this one is not equal to three. Okay. So it's not a palindrome. Okay. So that's what a palindrome is. Now, back to the problem, we're given a number K. So we're given the number of K and we want to find the smallest number palindrome, right? The smallest palindrome that is just larger than it, right? Just larger than K. So in this case K, we want to find the smallest palindrome just larger than K. So first input statement is 808. What is the smallest palindrome just larger than 808, right? So if you were to list out all the palindrome, the numbers from 808 and continue, so remember K is equal to 808 and just list out all the numbers 808, 809, 810, continue, continue. 811, 812, 813, 814, 815, 816, 817, 818, right? And you were to continually check if it's a palindrome or not. So 809 is not a palindrome, 810 is not a palindrome, 811 is not a palindrome, 812 is not a palindrome, right? So on so forth, we will figure out that 818 is a palindrome. And this is the smallest number, smallest palindrome just larger than 808, right? Because we only have took a few steps of adding one to get to 818. So now one way you could do this is literally do what we just said before is just keep adding one to this K value and then just checking, is it a palindrome? If it's a palindrome, then we just return true or false. Okay, so now the problem is that this takes too much time. So what is, how would you solve this problem? Okay, let's look at some really simple test cases to solve it. So let's say the number was 999, right? So what is the largest palindrome here? The smallest palindrome just larger than 99. Well if you keep adding one and one, you would get the smallest palindrome just larger than this is 101, okay? And if you were to check 999, the smallest palindrome just larger than this is going to be, if you keep adding one, you get 1000 at one, you get 1001. And if you check 9999, continue to add 1001, okay? So if all the numbers are nines, if all the numbers are nines, then that means that all you have to do is just literally just print out one, print out the number one. And then n minus one zeros and then print out another one, okay? So if it's all the values are nines, you just print out one and then the number of nines, the length of the value of k, the length of this k value. So then this k value initially, the length is equal to four, right? So if all the values are nines, right, you just have to print out one, then you print out three zeros, and then print out another one, okay? So this three is just, three is just the length of the number of nines minus one, okay? And in this case, it's the same thing. So the number of nines here is three, three, right? This length is three. So then in this case, we print out one, one, one, and then we print out two, two zeros, and then we print out one, one, right? So this is just going to be the length of the initial string of k minus one, which is two. So you print out one, one, and then two zeros and then a one, and then you do the same thing for this, nine, nine. So you print out one and then a zero and then a one, one, zero, okay? So that's the case of having all nines. And that's a simple case to figure out. But now let's go back to our values of eight, eight-oh-eight, okay? So k is equal to eight-oh-eight. How do I convert this number into the smallest palindrome of eight-eighteen? Well, if you just look at the number of eight-oh-eight and eight-eighteen, you realize that the only difference is the middle number of one, of zero and one. So if this is just the number of zero and one, all they did was they just added one to the middle number. This may work, this may work. But if you have a even number of numbers, let's say you have two, one, three, three. How are you going to, you can't just add one to two, one, three, three, right? What is the middle number? Right? The middle number, you don't know what the middle number is. So let's actually go to the test case of two, one, three, three and see how to do this problem. So remember k is equal to two, one, three, three and they said that the smallest palindrome of two, one, three, three, based on the test cases, right, is two, two, two, two. So if we were to go back to this, this value is two, two, two, two and this is the smallest palindrome. The smallest palindrome is larger than k. So how do they get two, two, two, okay? So thinking back to our initial idea of just adding one to the middle value and replacing it, let's actually look at two, one, three, three. But if we were to split this two, one, three, three into two parts, so we were to split this right in the middle, left side will be two, one and right side would be three, three. Now let's look at the right most digit of one and left most digit of three, three. So that in this case would just be with this one and this three. Now how did they get two, two, two, two, right? The smallest palindrome two, two, two, well, what did, what did they do? They took this one and they added it by one, right? Two, one, two, two is two, two, two, two is just two, one and added by one. So they took this number, they added it by one and they got two, two. And what did they do? They concatenated this two, two with its itself to get two, two, two, two, right? They took this last number, added by one, which gets you two, two and then they just concatenate it with itself to get two, two, two. So if the left most number is smaller than the right most number, the first digit of the right number, all you have to do is add one to the left most digit, which is two, two, and then just concatenate it with itself to get two, two, two, two, okay? So that's all you have to do. Okay, so that's one scenario. But what if you have a number that's larger than it? So let's say we have two, one, four, three. How would you do this problem, right? Well, oh no, not two, one, four, three, my fault. That's just the same thing as what we had before, two, four, three, seven, okay? So if we look at, if we split this number, okay, it's two, four, three, seven. And split this in half, two, four, and three, seven. So you have two, four, and then three, seven. Let's compare this first, these digits, four and three. Two, four, three, four is larger than three. So what do we do? Well, if you were to think about what is the smallest palindrome just larger than K? Two, four, three, seven, right? Can't we just print out the left most side and then reverse the digits, right? So from this point, we could just print out two, four, three, seven, right? Two, four, three, seven is our K value. The smallest palindrome just larger than K is actually two, four, four, two, okay? So can't we just take if this, if this left digit is greater than this right digit? Can't we just take our left digit and just print out itself, two, four, and concatenate it with its own digits reversed, four, two, right? Two, four, four, two. Because that would get you the smallest palindrome of two, four, four, two. So yeah, that's what you would do in order to do this problem. If it's greater than, if the four is greater than three, if the smallest five is greater than three, then we would actually do that. Now what if we get to a point where we have two there with equal? So let's say we have two, one. Let's go back to the case of, let's say we have eight, nine, seven, six. Okay, no, my fault, my fault. So let's say I go to a case where they're exactly equal. So let's say I have two, two, let's say I have like a one and then a seven and then like a six and a five or something like this, right? In this case, this two is equal to this two, okay? So what do I, what should I do at this point? Well at this point, you actually just continue going, right? You could just continue going, continue checking it. Because we don't know if we should concatenate it with itself or it with by itself. So we just actually keep going. So we're actually gonna loop down from this one to the left and loop this to the right. So at this point, we'll check this one and the seven, okay? So we continue going, check one and seven. Now, remember we split this in half, okay? Check one and seven. Now, is one less than seven? It is, one is less than seven. So what should we do? We take the leftmost digit of the left side of this, right? And add it by one. So we have six, one, two, right? Now we have six, one, three, right? Now, all we have to do is just concatenate it with itself and reverse the digits. So we would just have to do a three, one, six, okay? And that will be the smallest panel of the drone just larger than the initial value of here. So now let's actually explain the code and how we did this and then I'll be on our way. So let's go to the code right now and explain how to do this, okay? So initially we have our string of S. Now, let's actually check if they're all nines. Remember if they're all nines, all you have to do is print out one and then n minus one zeros and then print out one one. So if they're all nines, how do I check all nines? I just looped through the whole, I had a function called all nines. And I just looped through everything in the string. If it's not equal to a nine, I just return false. Otherwise, at the end, I'll return true. So then at this point, I check if all nines, if it's all nines, I print out one and I print out n minus one, which is the number of strings, the size of the string minus one, zeros, then I print out one one, right? And then I do a continue so that I could just go on to the next case. So I don't have to put an L statement, okay? All right, this is another case if they're all zeros. Now, I didn't go over that when you have all zeros because I'm not sure if the test case even has that. But if there's all zeros, what you would do is you would actually print out a one. N minus two zeros. So in this case, we have three minus two, which is one zero, and then we have a one. Wait, that would be your next palindrome. So that's what I did here. I check if they're all zeros, then I just print out one one. I print out n minus two zeros and I just print out a one, okay? So those are two specific test cases that are difficult to handle. Okay, now now it gets a little tricky. Okay, so I have the value of this. I create n is equal to the size of the string and I have a left and I have a right pointer. Initially, I set them both to be the middle value to be the middle number. So in this case, if we add like, I don't know, let's go back to the case of 808. Left and right pointer are both going to start in the middle. So this left and the right are both going to be in the middle. Okay, that's what I do here. All right. So remember the cases. So if, okay, yeah, in the middle. Now, if they are both, if the size mod two is even, now, if the number of letters are even number of the digits are even, you have to actually move left down one. And I could give you an example. Why? So back to 2143, right? The size of this number of digits, this is four. Right. So n is equal to four, the size of the digits here. Now, if I were to label the index is 0123, right? And if I set the left and right pointed to be n over two, so n is four. So four over two is two. So I set both of these at two. That's actually not right because I actually have to move the left down by one because if they're both at 0123, that's not the actual middle number. So we have two males, so we have to actually move left down by one. So I subtract one from left if it's, if the number of digits is even. So that's what I do here. So if number of digits is even, I subtract left by one. Okay. So now this is the part where it gets a little tricky. Remember back to our case when they're all the same in the middle. So if that's the case, remember we have two, five, two, two, seven, right? So in this case, left is equal to this part and the right corner is equal to this part. Now, if this, if they're equal, then I'm just going to move the left and right down by one. Okay. And I keep going that until I get to the end. So that's the reason why we'd have this. And we left down one and move the right down one. And that guarantees that at the end, I'm going to get, at least I get to a point where the left and the right are not the same. Okay. Now I get down to this part. Left pointer is less than zero or right, left pointer is less than right pointer. Okay. So in this case, if a left pointer is less than zero, then that means that I went past the string, right? So they're, they're definitely not equal at that point. They're definitely not equal. Now, if the left pointer is less than the right pointer, right? So this is just checking the last digit. Right. If the last digit is less than the very most last digit, right? So this is checking if I don't know, at this point, you're going to get five two to seven, you're going to get left pointer is five and the right pointer is going to equal to seven. And it's going to check five and seven. Okay. Now, if the left is less than the right, remember what we do is we have to add one to the left. So if the left is less than the right. Left is less than the right. So in this case, five is less than seven. We need to add one to left. So remember if I check if the left pointer, this part is going to be, if left point is greater than equal to zero. So how do I add one? Well, to add one, you need to for strings in this case, you actually have to continue going downward, right? And then you to add one to left, you have to continue going downward and then add a carry. So that's what this is doing, right? Cause in case if we have like a nine plus one, you get a zero, right? Nine plus one is 10, but you have to add the carry, right? So that's what this is for adding one. So in this code, what it's doing is that it's converting the left, that digit to it's the string of the left digit. So this five, it converts the character of five into its number format. So it would be five, right? That's what it's doing. Um, it's also adding a carry in case if you need to carry, right? So initially carry is one, then you add to carry. So the carry becomes six. Then what we do is we divide it by 10. So then we actually, uh, get the carry of the number. So in this case, it would take, uh, so in this case, remember, we have to add one to our left value, right? So remember if left is less than right, we add one to the left, add one to the left, right? So what we did was here, we added one to left digit, so make it six, right? Five plus one is six. Then what do we do? We divide it by 10, right? To get in case if we have a carry. So if it was a nine plus one, it was a 10, you have to carry the one, right? Or if it was like, uh, I don't know, 11 plus one, something like that. But, uh, yeah, you have to carry the one. So then we carry it, I take the number mod by 10, right? And then, um, here, this number mod by 10 is going to be the leftover value of the one's place, right? Left over value, the one's place. And then if you add by this, this, uh, this zero, what this does is it converts it into a character. So this is going to set the string to its corresponding value. So in this case, it would set the string to become six. Okay. Now you have to actually set the left pointer. The right pointer is also equal to left pointer. So this right, we're going to actually set it to six also. Okay. That's what it does. Then we move the left pointer and the right pointed down by one. So left pointer is going to move down to the next corresponding character. In this case, it would be empty. And the right is going to move down to the next, uh, corresponding character. So then now these are gone. Now we go to the next corresponding character. So that's what this is doing. Is just adding one to the left side and resetting the values to the right. Okay. Now, once we're done with that, let's say the right value was, uh, the left value was not less than the right value. So let's say it was greater. So the other case where Mario was like eight, uh, let's see, it was a eight, five, one, six, let's say, let's say we have this, right? We split in the middle. This five is greater than, than, uh, than the left side, right? This five is greater than the right side. Of one. So what do we do? Um, all we just have to do is just reset, flip the values of the right side to be the exact opposite of the values on the left. So in this case, the left pointer, our left point is pointing to here, right? And the right pointer is pointing to here. What we're going to do is we're going to set the right pointer to equal to left. So whatever character at the right pointer is going to equal to left. Our left is five. So at this right pointer of one, we're going to change this into five. Okay. And then we just continue going down, uh, then we moved on to the left, down one and the right, down one. So now the left, we moved down to here and then the right move down to here. Now is the right greater than the left? Uh, is the left greater than the right? Yeah, it is. So then we change the right to be left. Our left is eight, right. So right is now eight. And then that's it.  So yeah, that's basically how you do this, how you do this problem. Rate, comment, subscribe. Hope you guys enjoyed this video. I'll check you guys later. Peace.


Hi everyone, it's 11 o'clock on the east coast, we are the 21st of February 2018, and I'm Philippe LeFoucaret and I'm going to present to you the functional security update, functional group update for security products, sorry. This is the agenda for today, I will try to be brief because we have a lot of things to see. First of all, this is the new team for the security products, most of us are coming from the Genesium acquisition, so I'm Philippe LeFoucaret, the engineering manager. We have Fabio with us, which is with the interim product manager coming from the CIECD team and is helping us since the creation of this team. Fabio Encato is coming from Genesium as well. He was the leading developer at Genesium. We have Gilbert Roulot and Olivier Gonzalez with us. As you can see, it's a pretty senior team. Coming with some challenges, the first one is the transition from being synchronous to asynchronous. We had a very different way to solve things in Genesium, we were using the chat a lot, even if we were on different time zones. We used the chat to figure out things regarding issues and where everything was sorted out, we created the issue with the conclusion of the chat. It's a lot different in GitHub today, so we need to move to this asynchronous mode and it's pretty new for us and a big challenge. We also used to do CD, so continuous deployments to genesium.com. The way we used that was, again, a lot different from what GitLab is doing with the GitLab.com. Actually every change was pushed directly to master and then to production. We used genesium.com as a test lab for our enterprise customers. We did releases as well in Genesium, but genesium.com was evolving all the time and was a really great way to figure out issues before releasing things for Genesium Enterprise. Everyone is running at the same time, that means we can't or it's harder to ask questions to each other because we are on the same page on the onboarding process. The end book, as you know, is huge, it's more than 1,000 pages now or so. It's pretty much impossible to integrate the war end book in less than a month. I would like again to say a big shout out to Fabio because he's doing a lot to help us and he's everywhere and I know he's already doing a lot of things for the CI CD team. So again, thank you Fabio for the episode, I appreciate it. Just quickly for our objectives, as you know, we are migrating everything from Genesium to GitLab. So I won't go into details here, the idea is we want to make sure that the users from Genesium have the same experience in GitLab, so we are moving as much as possible. It's going pretty well because tomorrow we are going to have the 10.5 release of GitLab.com. And it includes a lot of things from Genesium already, so we are really happy with the accomplishment that we're going to see in the next slide. And also our objectives include making better security products that includes mostly SASD for now, but we're going to improve DAST as well. So far, we have succeeded in integrating the Genesium dependency checks into pipelines through the SASD job. That means for the user, a better Ruby support because we have mostly the same database as RubySec, but we have improved some adversaries and we have a little more adversaries than what they have. The same goes for NPM. We have a better support for NPM now. For Python, we have a better support as well because we are checking dependencies, which wasn't the case before the security products team integration. That's the big news. We have Java support in the SASD, so for now it's just dependencies checking, but that's a great step from what SASD was able to do. And again, it's an iteration, so that's the first iteration for Java. We're expecting for the next release 10.6 to have more than that. We want to have some static code analysis in SASD as well. And we also added a support for PHP through the Genesium because we're again checking dependencies there. We also added multi-languages detection in SASD. The job was stopping at the first language it was detecting. So now if we have a Ruby and NPM, for instance, in a project, both languages are going to be detected and managed by SASD. We have created a client and a server for SASD to use Genesium, so it's very specific to what we did in GitLab. It's already working and going to be shipped in 10.5. And we are removing the duplicates from the output because we're using many tools in the SASD and we might have some CVs several times in the repo, so that's a bit useless. We also updated the widget in the Merit Request. I won't take all the credits for that, I guess. The frontend team did everything. And it's really nice what we have now. As you can see, we had something with a lot of colors. It was a bit aggressive for the users and we have something very clean and we're working a lot with the UX team and the frontend team to improve that. And in the next weeks, you're going to see a lot of improvements there. And we have some great expectations for this widget and especially for the CI view, you're going to have that kind of report directly in the CI view, so for each commit. We have a few concerns as well since we joined GitLab. We have seen that GitLab does not use our tools. We have seen many projects using quote quality or RubyCop or the tools that we are using inside SASD, but in very custom configurations. So we need to figure out a way to make sure that everyone is using the SASD tools and the SASD jobs. And if it doesn't fit, we need to figure out why it doesn't fit and improve that. So that is going to be used in GitLab first because our users will have the same issues. We try to use SASD as well from the omnibus package, but unfortunately we can't do that because omnibus is using a lot more things than just packages dependencies. And we don't have that kind of data currently, but we really want to be able to secure everything inside GitLab and to enforce security requirements as soon as possible. And the last concern that we have is we don't have any metric. That means if we have that something in SASD, the image is going to be pushed directly to the registry and that image will be pulled for each new job. That means every time we change something, it's a kind of continuous deployment. So we're pushing directly to production and we have no feedback directly from the platform. We could have some feedback from the users, but that's all we're in. We shouldn't wait for the users to yell at us because we have break something. And that's it for today. Is there any questions? Could we talk about genesis and competitors and possibly explain the return or we compare? Right, that's a tough question. The main difference between what we're doing in GitLab and companies like checkmark is that we are leveraging some open source software to do SASD. That means we don't have anything in-house apart from Genesium to do SASD. So we're using code climate, we're using bundler audit, we're using Bandit for Python, etc. That's for me the biggest difference. And the company like checkmark, they are doing that internally. If you are using Codesc, for example, they are doing that internally. If you are using code climate, they have their own engine. So that's the main difference between the two. Do we have more questions? We're going to have a lot of presentations for non-engineering teams, especially cells. And so you will learn more of Genesium in the future. Yeah, I have great new t-shirts. It just arrived yesterday. That's a good timing. Okay, if, yeah. When I said I don't use these tools, can you elaborate? There are many, many, many projects inside GitLab, not only GitLab C and GitLab E. There is a SASD job in GitLab CE. I need to check that, really. But so far, on most projects that I've seen, we don't use SASD. And I don't want the developers to use SASD and to copy past the definition. I want the developer to be able to enable SASD. And in the future, we want to enable security requirements enforcement. That means Katie would be able to define, and Katie and our team would be able to define some requirements. And we should be able to enforce that on many projects with different levels. Okay, we don't have any more questions. I will give you five more seconds if you want to ask us something. And yes, if we're using autodevops, it's going to be enabled for every new project. But I suspect every new project is not using autodevops. I need to double-check that for the next FGU. After integrating into GL, are there any future plans? You can talk about future features. When you talk about future, I suspect you don't talk about gymnasium features. Yes, we have credit plans. But actually, what we want to do in the security products is to make the life of the developers easier. So we want to move that information that we have currently in the database to the eyes of the users. And most of the features will include reporting and dashboarding. Currently, we're just listing issues. And we want to make a lot more than that. Mac has made a great presentation, a great video a few days ago, about what code quality and SASD should be in the future. And we should really bring that information directly into the code. That's really what we want to be added. Yes, we have the auto update, but it's going to be on a very long term. I will leave that for the next FGU. Okay, thank you, everyone. I guess we have gone through all questions. Yeah, no more questions. So thank you again. And I will see you in the team call. Bye-bye.