1010. Pairs of Songs With Total Durations Divisible by 60.cpp

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
   Hide Hint #1
We only need to consider each song length modulo 60.
   Hide Hint #2
We can count the number of songs with (length % 60) equal to r, and store that in an array of size 60.







class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        vector<int> rem(60, 0);
        int res=0;

        for(auto &x: time){
            if(x%60==0) res+=rem[0];
            else res+=rem[60-x%60];

            rem[x%60]++;
        }

        return res;
    }
};