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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.

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

• 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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