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).

Copy

def find_it(seq):
    return [i for i in set(seq) if seq.count(i) % 2 == 1][0]
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