[Codility - Java] 2. Array - 2. OddOccurrencesInArray

OddOccurrencesInArray

Find value that occurs in odd number of elements.

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:

class Solution { public int solution(int[] 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.

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정수 N개로 이뤄진 배열 A
1 <= N < 1,000,000 이며
홀수개로 있는 숫자가 1개와, 나머지는 짝수개 있다.
홀수개가 있는 숫자를 반환하는 코드를 작성하면 된다.

간단하게, map을 이용하면 된다.
map에 key가 존재하면 remove하고,
없으면 put한다.

A 순회를 모두 마친 뒤 map에 남아있는 값이 바로 홀수개로 있는 정수다.

static int oddOccurrencesInArray(int[] A) {
    Map<Integer, Integer> map = new HashMap<>();
    for(int i : A) {
        if(map.containsKey(i)) {
            map.remove(i);
        } else {
            map.put(i, 1);
        }
    }
    return map.keySet().stream().findFirst().orElse(0);
}

$O(N)$ or $O(N*log(N))$