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# 37. 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`.

&#x20;

**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.
```

## Solution: (Two Sum using remainder)

```cpp
class Solution
{
public:
    int numPairsDivisibleBy60(vector<int> &time)
    {

        unordered_map<int, int> m;

        int res = 0;

        for (int i = 0; i < time.size(); i++)
        {
            int rem = time[i] % 60;

            if (rem == 0)
            {
                res += m[0];
            }
            else
            {
                int val = 60 - rem;
                if (m.find(val) != m.end())
                {
                    res += m[val];
                }
            }

            m[rem]++;
        }

        return res;
    }
};

```