Description
https://leetcode.com/problems/check-if-n-and-its-double-exist/
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
Explanation
Sort the list in order and check if the twice of the element value in the visited set.
Python Solution
class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
arr_in_order = sorted(arr)
visited = set()
for num in arr_in_order:
if num * 2 in visited:
return True
visited.add(num)
visited = set()
for num in arr_in_order[::-1]:
if num * 2 in visited:
return True
visited.add(num)
return False
- Time Complexity: O(N).
- Space Complexity: O(N).