pairs-of-songs-with-total-durations-divisible-by-60

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Algorithms

You are given a list of songs where the <code>ith</code> song has a duration of <code>time[i]</code> seconds.

Return the number of pairs of songs for which their total duration in seconds is divisible by <code>60</code>. Formally, we want the number of indices <code>i</code>, <code>j</code> such that <code>i &lt; j</code> with <code>(time[i] + time[j]) % 60 == 0</code>.

&nbsp;
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>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= time.length &lt;= 6 * 104</code></li>
	<li><code>1 &lt;= time[i] &lt;= 500</code></li>
</ul>

numPairsDivisibleBy60

Pairs of Songs With Total Durations Divisible by 60

Destroy Sequential Targets

destroy-sequential-targets

摧毁一系列目标

Count Pairs That Form a Complete Day II

count-pairs-that-form-a-complete-day-ii

构成整天的下标对数目 II

Array

Hash Table

Counting

在歌曲列表中，第 <code>i</code> 首歌曲的持续时间为 <code>time[i]</code> 秒。

返回其总持续时间（以秒为单位）可被 <code>60</code> 整除的歌曲对的数量。形式上，我们希望下标数字 <code>i</code> 和 <code>j</code> 满足&nbsp; <code>i &lt; j</code> 且有&nbsp;<code>(time[i] + time[j]) % 60 == 0</code>。

&nbsp;

示例 1：

<pre>
输入：time = [30,20,150,100,40]
输出：3
解释：这三对的总持续时间可被 60 整除：
(time[0] = 30, time[2] = 150): 总持续时间 180
(time[1] = 20, time[3] = 100): 总持续时间 120
(time[1] = 20, time[4] = 40): 总持续时间 60
</pre>

示例 2：

<pre>
输入：time = [60,60,60]
输出：3
解释：所有三对的总持续时间都是 120，可以被 60 整除。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= time.length &lt;= 6 * 104</code></li>
	<li><code>1 &lt;= time[i] &lt;= 500</code></li>
</ul>