Create 1013._Pairs_of_Songs_With_Total_Durations_Divisible_by_60.md

docs/Leetcode_Solutions/Python/1013._Pairs_of_Songs_With_Total_Durations_Divisible_by_60.md

+# 1013. Pairs of Songs With Total Durations Divisible by 60
+
+**<font color=red>难度: Easy</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/
+
+> 内容描述
+
+```
+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
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N^2)******- 空间复杂度: O(1)******
+
+先算出每种数字出现的次数
+
+然后取出2个数字看即可
+1. 两个数字不同
+2. 两个数字相同（但是必须保证这种数字出现至少2次）
+
+```python
+class Solution:
+    def numPairsDivisibleBy60(self, time):
+        s = collections.Counter(time)
+        uni_time = list(s.keys())
+        res = 0
+        for i in range(len(uni_time)): # 两个数字不同
+            for j in range(i+1, len(uni_time)):
+                if (uni_time[i] + uni_time[j]) % 60 == 0:
+                    res += s[uni_time[i]] * s[uni_time[j]]
+
+        for i in range(len(uni_time)): # 两个数字相同
+            if (uni_time[i] + uni_time[i]) % 60 == 0:
+                cnt = s[uni_time[i]]
+                if s[uni_time[i]] >= 2: # 但是必须保证这种数字出现至少2次
+                    res += self.permu(cnt) // (self.permu(2) * self.permu(cnt-2))
+        return res
+    
+    def permu(self, cnt):
+        res = 1
+        for i in range(1, cnt+1):
+            res *= i
+        return res
+```
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