Check If N and Its Double Exist LeetCode Solution

Problem – Check If N and Its Double Exist LeetCode Solution

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

Check If N and Its Double Exist LeetCode Solution in Java

public boolean checkIfExist(int[] arr) {
	HashSet<Integer> set = new HashSet<>();
	for(int a : arr) {
		if(set.contains(a*2) || (a%2 == 0 && set.contains(a/2))) return true;
		set.add(a);
	}
	return false;
}

Check If N and Its Double Exist LeetCode Solution in C++

class Solution {
public:
    bool checkIfExist(vector<int>& arr) {
        unordered_set<int> set;
        for(int i=0;i<arr.size();i++){
            if(set.count(2*arr[i])>0 || ((arr[i]%2==0) && set.count(arr[i]/2)>0))
                return true;
            set.insert(arr[i]);
        }
        return false;
    }
};

Check If N and Its Double Exist LeetCode Solution in Python

class Solution:
    def checkIfExist(self, arr: List[int]) -> bool:
        possible_M = set()
        possible_N = set()
        for a in arr:
            if a in possible_M: return True
            if a in possible_N: return True
            if a % 2 == 0: possible_M.add(a//2)
            possible_N.add(a*2)
        return False

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