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