Given an array, find the integer that appears an odd number of times.
There will always be only one integer that appears an odd number of times.
def find_it(seq):
pass
def find_it(seq):
return min([n for n in seq if seq.count(n) % 2 != 0])
def find_it(seq):
return [i for i in seq if seq.count(i) % 2 != 0][0]