# Given an array of integers, find the one that appears an odd number of times.
# There will always be only one integer that appears an odd number of times.
# Examples
# [7] should return 7, because it occurs 1 time (which is odd).
# [0] should return 0, because it occurs 1 time (which is odd).
# [1,1,2] should return 2, because it occurs 1 time (which is odd).
# [0,1,0,1,0] should return 0, because it occurs 3 times (which is odd).
# [1,2,2,3,3,3,4,3,3,3,2,2,1] should return 4, because it appears 1 time (which is odd).

def find_it(seq):
    for i in seq:
        if seq.count(i)%2!=0:
            return i


print(find_it([10]), 10);
print(find_it([20,1,1,2,2,3,3,5,5,4,20,4,5]), 5);
print(find_it([1,1,2,-2,5,2,4,4,-1,-2,5]), -1)
print(find_it([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5]), 5)