Pairs of Songs With Total Durations Divisible by 60 풀이

문제: https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/description/

Author: gringrape

20230107/pairs-of-songs/README.md

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+# Pairs of Songs With Total Durations Divisible by 60
+## 문제 분석
+### 구하는 것
+- pairs of songs
+### 주어진 것
+- list of songs
+- songs[i] => duration of the song
+### 조건
+- pair 로 묶인 songs 의 길이 합이 60으로 나누어떨어져야함.
+## 풀이 계획
+### 예시 분석
+[30, 20, 150, 100, 40]
+
+=> 
+[30, 150]
+[20, 100]
+[20, 40]
+
+=> 이렇게 탐색하는 경우 n^2의 시간복잡도를 가지게 됨. 
+
+60으로 먼저 나눠보면 어떨까?
+그 후에 나머지 배열을 순회하며 해시에 카운트를 저장하자. 
+처음에는 0, 짝이 되는 숫자가 등장하면 1 씩 증가.
+마지막에는 총 합을 구하면 된다.
+
+[30, 20, 30, 40, 40]
+{
+  30: 2,
+  20: 1,
+  40: 2,
+}
+
+[60, 60, 60] => [0, 0, 0]
+{
+  0: 3
+}
+
+

20230107/pairs-of-songs/python/test_numPairsDivisibleBy60.py

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+def numPairsDivisibleBy60(songs):
+    return count_pairs(counts=count_remainder(songs))
+
+
+def count_remainder(songs):
+    remainders = [song % 60 for song in songs]
+    counts = dict()
+    for remainder in remainders:
+        counts.setdefault(remainder, 0)
+        counts[remainder] += 1
+    return counts
+
+
+def test_count_remainder():
+    assert count_remainder([0, 10, 20, 10]) == {0: 1, 10: 2, 20: 1}
+
+
+def count_pairs(counts):
+    remainders = [0, 10, 20, 30]
+
+    count = 0
+    for remainder in remainders:
+        count_a = counts.get(remainder, 0)
+        count_b = counts.get(60 - remainder, 0)
+
+        match remainder:
+            case 0 | 30:
+                count += count_a * (count_a - 1) // 2
+            case _:
+                count += count_a * count_b
+
+    return count
+
+
+def test_count_pairs():
+    assert count_pairs({0: 2, 10: 2, 20: 3, 40: 5, 50: 3}) == 22
+    assert count_pairs({0: 3}) == 3
+
+
+def test_numPairsDivisibleBy60():
+    assert numPairsDivisibleBy60([30, 20, 150, 100, 40]) == 3
+    assert numPairsDivisibleBy60([60, 60, 60]) == 3