README_EN.md

1346. Check If N and Its Double Exist

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Description

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

Solutions

Python3

class Solution:
    def checkIfExist(self, arr: List[int]) -> bool:
        map = defaultdict(int)
        for i, num in enumerate(arr):
            map[num] = i
        for i, num in enumerate(arr):
            if num << 1 in map and i != map[num << 1]:
                return True
        return False

Java

class Solution {
    public boolean checkIfExist(int[] arr) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            map.put(arr[i], i);
        }
        for (int i = 0; i < arr.length; i++) {
            if (map.containsKey(arr[i] << 1) && i != map.get(arr[i] << 1))
                return true;
        }
        return false;
    }
}

TypeScript

function checkIfExist(arr: number[]): boolean {
    for (let i = arr.length - 1; i >= 0; --i) {
        let cur = arr[i];
        let t1 = 2 * cur;
        if (arr.includes(t1) && arr.indexOf(t1) != i) {
            return true;
        }
        let t2 = cur >> 1;
        if (cur % 2 == 0 && arr.includes(t2) && arr.indexOf(t2) != i) {
            return true;
        }
    }
    return false;
}

C++

class Solution {
public:
    bool checkIfExist(vector<int>& arr) {
        unordered_map<int, int> map;
        for (int i = 0; i < arr.size(); ++i) {
            map[arr[i]] = i;
        }
        for (int i = 0; i < arr.size(); ++i) {
            if (map.find(arr[i] * 2) != map.end() && i != map[arr[i] * 2]) {
                return true;
            }
        }
        return false;
    }
};

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