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# 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;
}
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)
return False
``````
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