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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Write a function
Step1:  designing algorithm
Step2:  version 2-
Step3:  V3
Step4:  V4
Step5:  V 5 - Some where long the lines forgot about the fine line "All but one occurs Even number of times". Instead solved it for "all but one int does not have any pair, is only a single element". Fixing the solutions now.
Step6:  V5 .1
Step7:  V 5_2
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Python Code:
A = [9,3,9,3,9,7,9]
print len(A)
print sorted(A)
Explanation: Write a function:
def solution(A)
that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.
For example, given array A such that:
  A[0] = 9  A[1] = 3  A[2] = 9
  A[3] = 3  A[4] = 9  A[5] = 7
  A[6] = 9
the function should return 7, as explained in the example above.
Assume that:
    N is an odd integer within the range [1..1,000,000];
    each element of array A is an integer within the range [1..1,000,000,000];
    all but one of the values in A occur an even number of times.
Complexity:
    expected worst-case time complexity is O(N);
    expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
Creating sample Array
End of explanation
single_num = sorted(A)[0]
count = 0
if len(A) == 0:
    print A
    
for num in sorted(A):
    print "num: ", num
    
    if num % single_num:
        print "single num: ", single_num
        
        print "count: ", count
        
        if count > 1:
            single_num = num
            count = 0
        else:
            print single_num
            
            break
    
    count += 1
   
#print single_num
def solution_1(A):
    
    if len(A) == 0:
        return A
    
    elif len(A) == 1:
        return A[0]
    
    elif type(A) is int:
        return A
    
    elif len(A) % 2:
        single_num = sorted(A)[0]
        
        count = 0
            
        for i,num in enumerate(sorted(A)):
            if num != single_num:
                print "num: ", num
                
                print "single num: ", single_num
                
                print "Count: ", count
                
                if count > 1:
                    single_num = num
                    count = 1
                elif count == 1:
                    return single_num
            else:
                count += 1
            if i >= len(A)-1:
                return single_num
            
            print "end single num: ", single_num
            
            print "end count: ", count
            
            print "end num: ", num
print sorted(A), "\n"
print solution(A)
B = [2, 2, 3, 3, 4]
print sorted(B), "\n"
print solution(B)
Explanation: designing algorithm
End of explanation
def solution_2(A):
    
    if len(A) == 0:
        return A
    
    elif len(A) == 1:
        return A[0]
    
    elif type(A) is int:
        return A
    
    elif len(A) % 2:
        single_num = sorted(A)[0]
        
        count = 0
            
        for i,num in enumerate(sorted(A)):
            if num != single_num: 
                print "num: ", num
                
                print "Count: ", count
                
                print "single num: ", single_num
                
                if count > 1:
                    single_num = num
                    
                    count = 1
                
                elif count == 1:
                    return str(single_num) +" " + str(sorted(A[i-5:i+5])) 
            else:
                count += 1
            
            if i >= len(A)-1:
                return single_num
print sorted(B)
print solution_2(B)
print sorted(A)
print solution_2(A)
Explanation: version 2-
End of explanation
def solution_3(A):
    
    if len(A) == 0:
        return A
    
    elif len(A) == 1:
        return A[0]
    
    elif type(A) is int:
        return A
    
    elif len(A) % 2:
        single_num = sorted(A)[0]
        
        count = 1
            
        for i,num in enumerate(sorted(A)):
            if num != single_num:    
                if count > 1:
                    single_num = num
                    
                    count = 1
                
                elif count == 1:
                    return single_num
            else:
                count += 1
            
            if i >= len(A)-1:
                return single_num
Explanation: V3
End of explanation
def solution_4(A):
    
    if len(A) == 0:
        return A
    
    elif len(A) == 1:
        return A[0]
    
    elif type(A) is int:
        return A
    
    elif len(A) % 2:
        single_num = sorted(A)[0]
        
        count = 1
            
        for i,num in enumerate(sorted(A)):
            if num != single_num:    
                if count > 1:
                    single_num = num
                    
                    count = 1
                
                elif count == 1:
                    return single_num
            elif single_num == num:
                count += 1
            
            if i >= len(A)-1:
                return single_num
sorted(A)
print A
print solution_4(A)
Explanation: V4
End of explanation
def solution_5(A):
    
    if len(A) == 0:
        return A
    
    elif len(A) == 1:
        return A[0]
    
    elif type(A) is int:
        return A
    
    elif len(A) % 2:
        single_num = sorted(A)[0]
        
        count = 0
            
        for i,num in enumerate(sorted(A)):
            if num != single_num:    
                if count > 1:
                    if count % 2:
                        print "count", count
                        
                        print "num", num
                        
                        print "single num", single_num
                        
                        return single_num
                    
                    single_num = num
                    
                    count = 1
                
                elif count == 1:
                    print "count", count
                    
                    print "num", num
                    
                    print "single num", single_num
                    
                    return single_num
            elif single_num == num:
                count += 1
            
            if i >= len(A)-1:
                return single_num
print solution_5(A)
1 % 2
Explanation: V 5 - Some where long the lines forgot about the fine line "All but one occurs Even number of times". Instead solved it for "all but one int does not have any pair, is only a single element". Fixing the solutions now.
End of explanation
def solution_5_1(A):
    
    if len(A) == 0:
        return A
    
    elif len(A) == 1:
        return A[0]
    
    elif type(A) is int:
        return A
    
    elif len(A) % 2:
        single_num = sorted(A)[0]
        
        count = 0
            
        for i,num in enumerate(sorted(A)):
            if num != single_num:    
                if count % 2:
                    print "count", count
                    
                    print "num", num
                    
                    print "single num", single_num
                    
                    return single_num
                
                single_num = num
                
                count = 1
                
            elif single_num == num:
                count += 1
            
            if i >= len(A)-1:
                return single_num
print solution_5_1(A)
Explanation: V5 .1
End of explanation
def solution_5_2(A):
    
    if len(A) == 0:
        return A
    
    elif len(A) == 1:
        return A[0]
    
    elif type(A) is int:
        return A
    
    elif len(A) % 2:
        single_num = sorted(A)[0]
        
        count = 0
            
        for i,num in enumerate(sorted(A)):
            if num != single_num:    
                if count % 2:            
                    return single_num
                else:
                    single_num = num
                
                    count = 1
                
            elif single_num == num:
                count += 1
            
            if i >= len(A)-1:   
                if count % 2:            
                    return single_num
                else
                    #just in case-
                    return None
print solution_5_2(A)
Explanation: V 5_2
End of explanation
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