1010-pairs-of-songs-with-total-durations-divisible-by-60.py

"""
Problem Link: https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/

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 
with (time[i] + time[j]) % 60 == 0.

Example 1:
Input: [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: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.
 
Note:
1 <= time.length <= 60000
1 <= time[i] <= 500
"""
class Solution(object):
    def numPairsDivisibleBy60(self, time):
        """
        :type time: List[int]
        :rtype: int
        """
        d = collections.Counter()
        count = 0
        for t in time:
            count += d[-t % 60]
            d[t % 60] += 1
        return count