1346. Check If N and Its Double Exist LeetCode Solution
In this guide, you will get 1346. Check If N and Its Double Exist LeetCode Solution with the best time and space complexity. The solution to Check If N and Its Double Exist problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of 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
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(n)
1346. Check If N and Its Double Exist LeetCode Solution in C++
class Solution {
public:
bool checkIfExist(vector<int>& arr) {
unordered_set<int> seen;
for (const int a : arr) {
if (seen.contains(a * 2) || a % 2 == 0 && seen.contains(a / 2))
return true;
seen.insert(a);
}
return false;
}
}; /* code provided by PROGIEZ */
1346. Check If N and Its Double Exist LeetCode Solution in Java
class Solution {
public boolean checkIfExist(int[] arr) {
Set<Integer> seen = new HashSet<>();
for (final int a : arr) {
if (seen.contains(a * 2) || a % 2 == 0 && seen.contains(a / 2))
return true;
seen.add(a);
}
return false;
}
} // code provided by PROGIEZ
1346. Check If N and Its Double Exist LeetCode Solution in Python
class Solution:
def checkIfExist(self, arr: list[int]) -> bool:
seen = set()
for a in arr:
if a * 2 in seen or a % 2 == 0 and a // 2 in seen:
return True
seen.add(a)
return False # code by PROGIEZ
