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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.

  • 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;
        }
    }
    
    ############
    
    class Solution {
        public boolean checkIfExist(int[] arr) {
            Map<Integer, Integer> m = new HashMap<>();
            int n = arr.length;
            for (int i = 0; i < n; ++i) {
                m.put(arr[i], i);
            }
            for (int i = 0; i < n; ++i) {
                if (m.containsKey(arr[i] << 1) && m.get(arr[i] << 1) != i) {
                    return true;
                }
            }
            return false;
        }
    }
    
  • // 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;
        }
    };
    
  • class Solution:
        def checkIfExist(self, arr: List[int]) -> bool:
            m = {v: i for i, v in enumerate(arr)}
            return any(v << 1 in m and m[v << 1] != i for i, v in enumerate(arr))
    
    
    
  • func checkIfExist(arr []int) bool {
    	m := make(map[int]int)
    	for i, v := range arr {
    		m[v] = i
    	}
    	for i, v := range arr {
    		if j, ok := m[v*2]; ok && j != i {
    			return true
    		}
    	}
    	return false
    }
    
  • function checkIfExist(arr: number[]): boolean {
        const s = new Set();
        for (const v of arr) {
            if (s.has(v << 1) || s.has(v / 2)) {
                return true;
            }
            s.add(v);
        }
        return false;
    }
    
    
  • /**
     * @param {number[]} arr
     * @return {boolean}
     */
    var checkIfExist = function (arr) {
        const s = new Set();
        for (const v of arr) {
            if (s.has(v << 1) || s.has(v / 2)) {
                return true;
            }
            s.add(v);
        }
        return false;
    };
    
    
  • use std::cmp::Ordering;
    impl Solution {
        pub fn check_if_exist(mut arr: Vec<i32>) -> bool {
            arr.sort();
            let n = arr.len();
            for i in 0..n {
                let target = arr[i] * 2;
                let mut left = 0;
                let mut right = n;
                while left < right {
                    let mid = left + (right - left) / 2;
                    match arr[mid].cmp(&target) {
                        Ordering::Less => left = mid + 1,
                        Ordering::Greater => right = mid,
                        Ordering::Equal => {
                            if mid == i {
                                break;
                            }
                            return true;
                        }
                    }
                }
            }
            false
        }
    }
    
  • class Solution {
        /**
         * @param Integer[] $arr
         * @return Boolean
         */
        function checkIfExist($arr) {
            for ($i = 0; $i < count($arr); $i++) {
                $hashtable[$arr[$i] * 2] = $i;
            }
            for ($i = 0; $i < count($arr); $i++) {
                if (isset($hashtable[$arr[$i]]) && $hashtable[$arr[$i]] != $i) return true;
            }
            return false;
        }
    }
    

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