## Algorithm

Problem Name: 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 = 30, time = 150): total duration 180
(time = 20, time = 100): total duration 120
(time = 20, time = 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`

## Code Examples

### #1 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
class Solution:
def numPairsDivisibleBy60(self, time: List[int]) -> int:
mod =  * 61
for t in time:
mod[-1] += mod[(60 - t % 60) % 60]
mod[t % 60] += 1
return mod[-1]
``````
Copy The Code &

Input

cmd
time = [30,20,150,100,40]

Output

cmd
3

### #2 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _1010_PairsOfSongsWithTotalDurationsDivisibleBy60
{
public int NumPairsDivisibleBy60(int[] time)
{
var counts = new int;
var result = 0;
foreach (var num in time)
{
result += counts[(600 - num) % 60];
counts[num % 60]++;
}

return result;
}
}
}
``````
Copy The Code &

Input

cmd
time = [30,20,150,100,40]

Output

cmd
3