Pairs of songs with total durations divisible by 60

Time: O(N); Space: O(1); easy

In a list of songs, 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.

Notes:

  • 1 <= len(time) <= 60000

  • 1 <= time[i] <= 500

[1]:

import collections

class Solution1(object):
    def numPairsDivisibleBy60(self, time):
        """
        :type time: List[int]
        :rtype: int
        """
        result = 0
        count = collections.Counter()
        for t in time:
            result += count[-t % 60]
            count[t % 60] += 1
        return result

[2]:

s = Solution1()
time = [30, 20, 150, 100, 40]
assert s.numPairsDivisibleBy60(time) == 3
time = [60, 60, 60]
assert s.numPairsDivisibleBy60(time) == 3