Formatted question description: https://leetcode.ca/all/1346.html

1346. Check If N and Its Double Exist (Easy)

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

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/check-if-n-and-its-double-exist/

// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool checkIfExist(vector<int>& arr) {
        unordered_set<int> s;
        for (int n : arr) {
            if (s.count(2 * n) || (n % 2 == 0 && s.count(n / 2))) return true;
            s.insert(n);
        }
        return false;
    }
};

Java

class Solution {
    public boolean checkIfExist(int[] arr) {
        Set<Integer> set = new HashSet<Integer>();
        for (int num : arr) {
            if (num != 0)
                set.add(num);
            else {
                if (!set.add(num))
                    return true;
            }
        }
        for (int num : arr) {
            if (num != 0 && set.contains(num * 2))
                return true;
        }
        return false;
    }
}

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