[코딜리티] Lesson 2 - Arrays : OddOccurrencesInArray

Lesson 2 - Arrays : OddOccurrencesInArray

📌 문제

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.

📝 풀이

• 입력받은 배열을 반복순회하며 자료구조에 담아준다.
• 이때, 자료구조에 이미 있다면 해당 요소은 넣지않고 있던 요소도 제거해준다.
• 배열을 모두 순회한 뒤 자료구조에 남아있는 값을 출력

💻 구현

static class Solution {
    public int solution(int[] A) {
        Set<Integer> set = new HashSet<>();
        for(int i = 0; i < A.length; i++){
            if(set.contains(A[i])){
                set.remove(A[i]);
            } else {
                set.add(A[i]);
            }
        }
        return set.stream().iterator().next();
    }
}