1010. Pairs of Songs With Total Durations Divisible by 60

You are given a list of songs where the ith song has a duration of time[i] seconds.

Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0.

 

Example 1:

Input: time = [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60

Example 2:

Input: time = [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

 

Constraints:

  • 1 <= time.length <= 6 * 104
  • 1 <= time[i] <= 500

1010. Pairs of Songs With Total Durations Divisible by 60
struct Solution;

impl Solution {
    fn num_pairs_divisible_by60(time: Vec<i32>) -> i32 {
        let mut a: Vec<i32> = vec![0; 60];
        let mut res = 0;
        for x in time {
            let count = a[((600 - x) % 60) as usize];
            if count != 0 {
                res += count;
            }
            a[(x % 60) as usize] += 1;
        }
        res
    }
}

#[test]
fn test() {
    let time = vec![30, 20, 150, 100, 40];
    assert_eq!(Solution::num_pairs_divisible_by60(time), 3);
    let time = vec![60, 60, 60];
    assert_eq!(Solution::num_pairs_divisible_by60(time), 3);
}