Building an Aquarium
You love fish, that's why you have decided to build an aquarium. You have a piece of coral made of $$$n$$$ columns, the $$$i$$$-th of which is $$$a_i$$$ units tall. Afterwards, you will build a tank around the coral as follows:   Pick an integer $$$h \geq 1$$$ — the height of the tank. Build walls of height $$$h$$$ on either side of the tank.  Then, fill the tank up with water so that the height of each column is $$$h$$$, unless the coral is taller than $$$h$$$; then no water should be added to this column.  For example, with $$$a=[3,1,2,4,6,2,5]$$$ and a height of $$$h=4$$$, you will end up using a total of $$$w=8$$$ units of water, as shown.    You can use at most $$$x$$$ units of water to fill up the tank, but you want to build the biggest tank possible. What is the largest value of $$$h$$$ you can select?
The first test case is pictured in the statement. With $$$h=4$$$ we need $$$8$$$ units of water, but if $$$h$$$ is increased to $$$5$$$ we need $$$13$$$ units of water, which is more than $$$x=9$$$. So $$$h=4$$$ is optimal.In the second test case, we can pick $$$h=4$$$ and add $$$3$$$ units to each column, using a total of $$$9$$$ units of water. It can be shown that this is optimal.In the third test case, we can pick $$$h=2$$$ and use all of our water, so it is optimal.
The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.The first line of each test case contains two positive integers $$$n$$$ and $$$x$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq x \leq 10^9$$$) — the number of columns of the coral and the maximum amount of water you can use.The second line of each test case contains $$$n$$$ space-separated integers $$$a_i$$$ ($$$1 \leq a_i \leq 10^9$$$) — the heights of the coral.The sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
For each test case, output a single positive integer $$$h$$$ ($$$h \geq 1$$$) — the maximum height the tank can have, so you need at most $$$x$$$ units of water to fill up the tank.We have a proof that under these constraints, such a value of $$$h$$$ always exists.
def main():
    t = int(input())
    r = [str(solution(i, t)) for i in range(t)]
    print('\n'.join(r))

def solution(ti, tn):
    (n, x) = map(int, input().split())
    aa = list(map(int, input().split()))
    aa.sort()
    i = 1
    while i < n and 0 < x:
        x -= i * (aa[i] - aa[i - 1])
        i += 1
    return aa[i - 1] + x // (i if x > 0 else i - 1)
main()
def main():
    t = int(input())
    r = [str(solution(i, t)) for i in range(t)]
    print('\n'.join(r))

def solution(ti, tn):
    (n, x) = map(int, input().split())
    aa = list(map(int, input().split()))
    aa.sort()
    for i in range(1, n):
        d = aa[i] - aa[i - 1]
        s = d * i
        x -= s
        if x == 0:
            return aa[i]
        if x < 0:
            return aa[i - 1] + (x + s) // i
    if x == 0:
        return aa[n - 1]
    if x > 0:
        return aa[n - 1] + x // n
main()
import sys
infile = sys.stdin
input = lambda : infile.readline().rstrip()
for _ in range(int(input())):
    (n, x) = map(int, input().split())
    (*la,) = map(int, input().split())
    k = min(la)
    m = max(la)
    score = m * n - sum(la)
    if score <= x:
        print(m + (x - score) // n)
    else:
        sla = sorted(la)
        score = 0
        sumI = 0
        for (i, va) in enumerate(sla):
            sumI += va
            vol = va * (i + 1) - sumI
            if vol >= x:
                deltaSup = vol - x
                (q, r) = divmod(deltaSup, i)
                if r:
                    q += 1
                print(va - q)
                break