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

/*
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

Hint #1  
We only need to consider each song length modulo 60.

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);
    for (auto &e : time)
      rem[e % 60]++;

    int ans = 0;
    if (rem[0])
      ans += ((rem[0]) * (rem[0] - 1) / 2); // nC2
    if (rem[30])
      ans += ((rem[30]) * (rem[30] - 1) / 2);

    for (int i = 1; i <= 29; ++i)
      ans += rem[i] * rem[60 - i];

    return ans;
  }
};