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# QUESTION
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# 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.

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# SOLUTION
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def find_it(seq):
    x = list(dict.fromkeys(seq))

    u = []
    for w in x:
        i = 0
        for k in seq:
            if k == w:
                i += 1
        u.append(i)

    for k in list(range(0,len(u))):
        if u[k] % 2 != 0:
            return x[k]
