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You are given a list of songs where the i<sup>th</sup> 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:
<pre>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 </pre>
Example 2:
<pre>Input: time = [60,60,60] Output: 3 Explanation: All three pairs have a total duration of 120, which is divisible by 60. </pre>
Constraints:
1 <= time.length <= 6 * 10<sup>4</sup>1 <= time[i] <= 500
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题目大意：
求数组中两数和能被60整除
解题思路：
两数和的关系第一时间想到two sum。但由于target是60的倍数，并不固定，所以先用公式求所有数的mod，(time[i] + time[j]) % 60 = time[i] % 60 + time[j] % 60, 这样target就是60了
第二个难点是此题求个数并不是像two sum一样求可行性，所以value to index改成value to count
解题步骤：
N/A
注意事项：
- 对所有数对60求mod，map存value到count
- 如果输入是60，取模后为0， 求(60 - time_mod[i])要对60取模，否则60不在map中，因为60 - time_mod[i] = 60.
Python代码：
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10# (time[i] + time[j]) % 60 = time[i] % 60 + time[j] % 60
def numPairsDivisibleBy60(self, time: List[int]) -> int:
time_mod = [t % 60 for t in time] # [30,30]
val_to_count = collections.defaultdict(int)
res = 0
for i in range(len(time_mod)):
if (60 - time_mod[i]) % 60 in val_to_count:
res += val_to_count[(60 - time_mod[i]) % 60]
val_to_count[time_mod[i]] += 1 # 30:1
return res
算法分析：
时间复杂度为O(n)，空间复杂度O(n)
