Solution 1 - Hash Set (One Pass)

For each x, check if 2x or x/2 (when even) already seen.

TimeComplexity: O(n)

SpaceComplexity: O(n)

""" 1346.1 - Check If N and Its Double Exist - Solution 1 - Hash Set """

from typing import List

class Solution:
    def checkIfExist(self, arr: List[int]) -> bool:
        seen = set()
        for x in arr:
            if 2*x in seen or (x % 2 == 0 and x//2 in seen):
                return True
            seen.add(x)
        return False

function checkIfExist(arr:number[]):boolean{
 const seen:Set<number>=new Set();
 for(const x of arr){
  if(seen.has(2*x) || (x%2===0 && seen.has(x/2))) return true;
  seen.add(x);
 }
 return false;
}

package main

func checkIfExist(arr []int) bool {
	seen := map[int]bool{}
	for _, x := range arr {
		if seen[2*x] || (x%2 == 0 && seen[x/2]) {
			return true
		}
		seen[x] = true
	}
	return false
}

func main() {}

use std::collections::HashSet;

fn check_if_exist(arr: &[i32]) -> bool {
	let mut seen: HashSet<i32> = HashSet::new();
	for &x in arr {
		if seen.contains(&(2*x)) || (x%2==0 && seen.contains(&(x/2))) { return true; }
		seen.insert(x);
	}
	false
}
fn main() {}

1346 - Check If N and Its Double Exist

Solution 1 - Hash Set (One Pass)