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

Solution:

count the frequencies of remainders,
if we have 2 numbers with remainders 2, and 2 numbers with remainder 58, then we have 2 * 2 = 4 pairs which sum to 60
remainders[30] and remainders[0] are special cases because we can pair any two pairs inside of them

class Solution {
    public int numPairsDivisibleBy60(int[] time) {
        int[] count = new int[60];   
        for (int t : time) {
            count[t % 60] ++;
        }
        int res = count[0] * (count[0] - 1) / 2;
        for (int i = 1; i < 30; i ++) {
            res += count[i] * count[60 - i];
        }
        res += count[30] * (count[30] - 1) / 2;
        return res;
    }
}