Pairs of songs with total durations divisible by 60

class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        count = collections.defaultdict(int)
        res = 0
        for i in time:
            target = 60 - (i % 60)
            if i % 60 == 0:
                res += count[0]
            else:
                res += count[target]
            count[i % 60] += 1
        return res

Pairs of Songs With Total Durations Divisible by 60

You are given a list of songs where the i^th 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