1010 Pairs of Songs With Total Durations Divisible by 60.py

Author: rajeevranjancom

1010 Pairs of Songs With Total Durations Divisible by 60.py

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+#!/usr/bin/python3
+"""
+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
+"""
+from typing import List
+from collections import defaultdict
+
+
+class Solution:
+    def numPairsDivisibleBy60(self, time: List[int]) -> int:
+        """
+        Running attribution
+        """
+        counter = defaultdict(int)
+        ret = 0
+        for t in time:
+            ret += counter[(60 - t) % 60]  # handle 0
+            counter[t % 60] += 1
+
+        return ret
+
+
+    def numPairsDivisibleBy60_error(self, time: List[int]) -> int:
+        """
+        O(N^2) check i, j
+        Reduce it
+        O(N) by using hashmap, the key should mod 60
+
+        attribution error
+        """
+        hm = defaultdict(int)
+        for t in time:
+            hm[t % 60] += 1
+
+        ret = 0
+        for k, v in hm.items():
+            if k == 0:
+                ret += (v * (v - 1)) // 2
+            elif k <= 60 - k:  # attribution
+                v2 =  hm[60 - k]
+                ret += v2 * v
+
+        return ret