Check If N and Its Double Exist

Problem Id: 1346 Difficulty: Easy Tag: Array


Just use a hash set to solve this problem. This would use time complexity O(n) and space complexity O(n).

Pay attaintion to 0, because it is the same with its double.


class Solution:
    def checkIfExist(self, arr: List[int]) -> bool:
        tmp = set(arr)
        if 0 in tmp and arr.count(0) > 1:
            return True
        for num in arr:
            if num != 0 and 2 * num in tmp:
                return True
        return False