Pairs of Songs With Total Durations Divisible by 60 (python)

题目：

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.

 

解题思路：

要找到数组中的times[i]+times[j](i

代码如下：

class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        mod_times = {}
        for i in time:
            mod_times[i%60] = 1 if i%60 not in mod_times.keys() else mod_times[i%60]+1
        cnt = 0
        
        for key,value in mod_times.items():
            print(key,value)
            if key == 0 or key == 30:
                cnt += int((mod_times[key]*(mod_times[key]-1))/2)
            elif key < 30 and (60-key) in mod_times.keys():
                cnt += mod_times[key] * mod_times[60-key]
             
        return cnt