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.

Input:time = [30,20,150,100,40]Output:3Explanation: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

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

- 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; } }