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

2024-06-07 12:20:12

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

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

题目描述

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

C++ 实现 1

参看 [Java/C++/Python] Two Sum with K = 60, 思路非常详细. 大致如下:

将问题转化为 1. Two Sum*, 目标是对于 time 中的每个元素 t % 60, 在哈希表中查询是否存在 60 - (t % 60). 毕竟我们希望找到 x 使得 (x + t) % 60 == 0. 然而这里存在一个问题, 由于 t % 60 结果是 0 ~ 59, 但是 60 - (t % 60) 范围却是 1 ~ 60. 有两种思路:

判断 (60 - (t % 60)) % 60) 在哈希表中存在
判断 (600 - t) % 60 是否在哈希表中

class Solution {public:int numPairsDivisibleBy60(vector<int>& time) {unordered_map<int, int> records;int res = 0;for (auto &t : time) {res += records[(600 - t) % 60];records[t % 60] ++;}return res;}
};

C++ 实现 2

class Solution {public:int numPairsDivisibleBy60(vector<int>& time) {unordered_map<int, int> records;int res = 0;for (auto &t : time) {res += records[(60 - t % 60) % 60];records[t % 60] ++;}return res;}
};

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