Given an integer num, find the closest two integers in absolute difference whose product equals num + 1 or num + 2.
Return the two integers in any order.
Example 1:
Input: num = 8 Output: [3,3] Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.
Example 2:
Input: num = 123 Output: [5,25]
Example 3:
Input: num = 999 Output: [40,25]
Constraints:
1 <= num <= 10^9struct Solution;
impl Solution {
fn closest_divisors(num: i32) -> Vec<i32> {
for i in (0..=((num + 2) as f64).sqrt() as i32).rev() {
if (num + 1) % i == 0 {
return vec![(num + 1) / i, i];
}
if (num + 2) % i == 0 {
return vec![(num + 2) / i, i];
}
}
vec![]
}
}
#[test]
fn test() {
let num = 8;
let res = vec![3, 3];
assert_eq!(Solution::closest_divisors(num), res);
let num = 123;
let res = vec![25, 5];
assert_eq!(Solution::closest_divisors(num), res);
let num = 999;
let res = vec![40, 25];
assert_eq!(Solution::closest_divisors(num), res);
}