README.md

1346. 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

Solutions (Rust)

1. Set

use std::collections::HashSet;implSolution{pubfncheck_if_exist(arr:Vec<i32>) -> bool{letmut set = HashSet::new();for n in arr {if set.contains(&(2* n)) || (n % 2 == 0 && set.contains(&(n / 2))){returntrue;} set.insert(n);}false}}