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leetcode-problemset/leetcode-cn/problem (English)/节流(English) [throttle].html
2023-05-15 17:43:00 +08:00

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<p>Given a function <code>fn</code> and&nbsp;a time in milliseconds <code>t</code>, return&nbsp;a <strong>throttled</strong> version of that function.</p>
<p>A <strong>throttled</strong> function is first called without delay and then, for a time interval of <code>t</code> milliseconds, can&#39;t be executed but should store the latest function arguments provided to call <code>fn</code> with them after the end of the delay.</p>
<p>For instance, <code>t = 50ms</code>, and the function was called at <code>30ms</code>, <code>40ms</code>, and <code>60ms</code>. The first function call would block calling functions for the following <code>t</code> milliseconds. The second function call would save arguments, and the third call arguments should overwrite currently stored arguments from the second call because the second and third calls are called before <code>80ms</code>. Once the delay has passed, the throttled function should be called with the latest arguments provided during the delay period, and it should also create another delay period of <code>80ms + t</code>.</p>
<p><img alt="Throttle Diagram" src="https://assets.leetcode.com/uploads/2023/04/08/screen-shot-2023-04-08-at-120313-pm.png" style="width: 1156px; height: 372px;" />The above diagram&nbsp;shows how throttle&nbsp;will transform&nbsp;events. Each rectangle represents 100ms and the throttle&nbsp;time is 400ms. Each color represents a different set of inputs.</p>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> t = 100, calls = [{&quot;t&quot;:20,&quot;inputs&quot;:[1]}]
<strong>Output:</strong> [{&quot;t&quot;:20,&quot;inputs&quot;:[1]}]
<strong>Explanation:</strong> The 1st call is always called without delay
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> t = 50, calls = [{&quot;t&quot;:50,&quot;inputs&quot;:[1]},{&quot;t&quot;:75,&quot;inputs&quot;:[2]}]
<strong>Output:</strong> [{&quot;t&quot;:50,&quot;inputs&quot;:[1]},{&quot;t&quot;:100,&quot;inputs&quot;:[2]}]
<strong>Explanation:</strong>
The 1st is called a function with arguments (1) without delay.
The 2nd is called at 75ms, within the delay period because 50ms + 50ms = 100ms, so the next call can be reached at 100ms. Therefore, we save arguments from the 2nd call to use them at the callback of the 1st call.
</pre>
<p><strong class="example">Example 3:</strong></p>
<pre>
<strong>Input:</strong> t = 70, calls = [{&quot;t&quot;:50,&quot;inputs&quot;:[1]},{&quot;t&quot;:75,&quot;inputs&quot;:[2]},{&quot;t&quot;:90,&quot;inputs&quot;:[8]},{&quot;t&quot;: 140, &quot;inputs&quot;:[5,7]},{&quot;t&quot;: 300, &quot;inputs&quot;: [9,4]}]
<strong>Output:</strong> [{&quot;t&quot;:50,&quot;inputs&quot;:[1]},{&quot;t&quot;:120,&quot;inputs&quot;:[8]},{&quot;t&quot;:190,&quot;inputs&quot;:[5,7]},{&quot;t&quot;:300,&quot;inputs&quot;:[9,4]}]
<strong>Explanation:</strong>
The 1st is called a function with arguments (1) without delay.
The 2nd is called at 75ms within the delay period because 50ms + 70ms = 120ms, so it should only save arguments.&nbsp;
The 3rd is also called within the delay period, and because we need just the latest function arguments, we overwrite previous ones. After the delay period, we do a callback at 120ms with saved arguments. That callback makes another delay period of 120ms + 70ms = 190ms so that the next function can be called at 190ms.
The 4th is called at 140ms in the delay period, so it should be called as a callback at 190ms. That will create another delay period of 190ms + 70ms = 260ms.
The 5th is called at 300ms, but it is after 260ms, so it should be called immediately and should create another delay period of 300ms + 70ms = 370ms.</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>0 &lt;= t &lt;= 1000</code></li>
<li><code>1 &lt;= calls.length &lt;= 10</code></li>
<li><code>0 &lt;= calls[i].t &lt;= 1000</code></li>
<li><code>0 &lt;= calls[i].inputs[i], calls[i].inputs.length &lt;= 10</code></li>
</ul>