[Leetcode] Check If N and Its Double Exist ★★

Given an array arr of integers, check if there exists two integers N and M such that N is the double of M ( i.e. N = 2 * M).

More formally check if there exists two indices i and j such that :

• i != j
• 0 <= i, j < arr.length
• arr[i] == 2 * arr[j]

 

Example 1:

Input: arr = [10,2,5,3]

Output: true

Explanation: N = 10 is the double of M = 5,that is, 10 = 2 * 5.

Example 2:

Input: arr = [7,1,14,11]

Output: true

Explanation: N = 14 is the double of M = 7,that is, 14 = 2 * 7.

Example 3:

Input: arr = [3,1,7,11]

Output: false

Explanation: In this case does not exist N and M, such that N = 2 * M.

 

Constraints:

• 2 <= arr.length <= 500
• -10^3 <= arr[i] <= 10^3

class Solution {
public:
    bool checkIfExist(vector<int>& arr)
    {
        size_t size = arr.size();

        unordered_set<int> s;

        for (size_t i = 0; i < size; i++)
        {
            int num = arr[i] * 2;

            if (s.find(num) != s.end())
                return true;

            if (arr[i] % 2 == 0)
            {
                if (s.find(arr[i] / 2) != s.end())
                    return true;
            }

            s.insert(arr[i]);
        }

        return false;
    }
};

unordered_set : set과 달리 정렬되지 않고, Hash 함수 기반의 set