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1346. Check If N and Its Double Exist

Description

Given an array arr of integers, check if there exist 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: For i = 0 and j = 2, arr[i] == 10 == 2 * 5 == 2 * arr[j]

Example 2:

Input: arr = [3,1,7,11]
Output: false
Explanation: There is no i and j that satisfy the conditions.

 

Constraints:

• 2 <= arr.length <= 500
• -103 <= arr[i] <= 103

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• Php
• RenderScript

• 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;
      }
  }
  

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

• 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;
      }
  }
  
  

• use std::collections::HashMap;
  impl Solution {
      pub fn check_if_exist(arr: Vec<i32>) -> bool {
          let mut map = HashMap::new();
          for (i, v) in arr.iter().enumerate() {
              map.insert(v, i);
          }
          for (i, v) in arr.iter().enumerate() {
              if map.contains_key(&(v * 2)) && map[&(v * 2)] != i {
                  return true;
              }
          }
          false
      }
  }
  
  

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