1346 - Check If N and Its Double Exist

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

C++

• 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;
      }
  };

JAVA

• 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;
      }
  }

PYTHON

• class Solution:
      def checkIfExist(self, arr):
          m = {}
          n = len(arr)
          for i in range(n):
              m[arr[i]] = i
          for i in range(n):
              if arr[i] * 2 in m and m[arr[i] * 2] != i:
                  return True
          return False