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leetcode-problemset/leetcode-cn/problem (English)/爱生气的书店老板(English) [grumpy-bookstore-owner].html

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<p>There is a bookstore owner that has a store open for <code>n</code> minutes. Every minute, some number of customers enter the store. You are given an integer array <code>customers</code> of length <code>n</code> where <code>customers[i]</code> is the number of the customer that enters the store at the start of the <code>i<sup>th</sup></code> minute and all those customers leave after the end of that minute.</p>
<p>On some minutes, the bookstore owner is grumpy. You are given a binary array grumpy where <code>grumpy[i]</code> is <code>1</code> if the bookstore owner is grumpy during the <code>i<sup>th</sup></code> minute, and is <code>0</code> otherwise.</p>
<p>When the bookstore owner is grumpy, the customers of that minute are not satisfied, otherwise, they are satisfied.</p>
<p>The bookstore owner knows a secret technique to keep themselves not grumpy for <code>minutes</code> consecutive minutes, but can only use it once.</p>
<p>Return <em>the maximum number of customers that can be satisfied throughout the day</em>.</p>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], minutes = 3
<strong>Output:</strong> 16
<strong>Explanation:</strong> The bookstore owner keeps themselves not grumpy for the last 3 minutes.
The maximum number of customers that can be satisfied = 1 + 1 + 1 + 1 + 7 + 5 = 16.
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> customers = [1], grumpy = [0], minutes = 1
<strong>Output:</strong> 1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>n == customers.length == grumpy.length</code></li>
<li><code>1 &lt;= minutes &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
<li><code>0 &lt;= customers[i] &lt;= 1000</code></li>
<li><code>grumpy[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>