// 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

//

// Runtime 2 ms Beats 100%

// Memory 53.4 MB Beats 14.57%

class Solution {

public int numPairsDivisibleBy60(int[] time) {

long[] counts = new long[60];

for (int song: time) {

int mod = song % 60;

counts[mod]++;

}

long count = 0;

for (int i = 1; i < 30; i++) {

count = count + counts[i] * counts[60 - i];

}

count = count + counts[0] * (counts[0] - 1) / 2 + counts[30] * (counts[30] - 1) / 2;

return (int)count;

}

}