Find the odd int

Description:

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).

 

 

Solution:

1. Count the number of identical elements.
2. If the number of elements is odd, the element is returned.

 

 

def find_it(seq):
    for i in set(seq):
        if seq.count(i)&1:
            return i

 

 

How to solve using the collections built-in function.

import collections
        
def find_it(seq):
    for k, v in collections.Counter(seq).items():
        if v&1:
            return k