OddOccurrencesInArray

Link: https://app.codility.com/programmers/lessons/2-arrays/odd_occurrences_in_array/

A non-empty array A consisting of N integers is given.
The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.

For example, in array A such that:

A[0] = 9 A[1] = 3 A[2] = 9.
A[3] = 3 A[4] = 9 A[5] = 7.
A[6] = 9.
the elements at indexes 0 and 2 have value 9, the elements at indexes 1 and 3 have value 3, the elements at indexes 4 and 6 have value 9, the element at index 5 has value 7 and is unpaired. Write a function:

def solution(A)

that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.

For example, given array A such that:

A[0] = 9 A[1] = 3 A[2] = 9 A[3] = 3 A[4] = 9 A[5] = 7 A[6] = 9 the function should return 7, as explained in the example above.

Write an efficient algorithm for the following assumptions:

N is an odd integer within the range [1..1,000,000]; each element of array A is an integer within the range [1..1,000,000,000]; all but one of the values in A occur an even number of times.

Tried

처음에 문제를 잘못 이해했다. A[i+2] == A[i]를 비교하는 문제인 줄 알았는데,
보니까 각 숫자마다 짝수로 이루어져있고, 홀수로 이루어진 수를 찾는 문제였다..

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from collections import Counter

def solution(A):
    dict_A = Counter(A)
    for key in dict_A.keys():
        if dict_A[key] % 2 == 1:
            return key