grumpy-bookstore-owner

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Algorithms

There is a bookstore owner that has a store open for <code>n</code> minutes. You are given an integer array <code>customers</code> of length <code>n</code> where <code>customers[i]</code> is the number of the customers that enter the store at the start of the <code>ith</code> minute and all those customers leave after the end of that minute.

During certain 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>ith</code> minute, and is <code>0</code> otherwise.

When the bookstore owner is grumpy, the customers entering during that minute are not satisfied. Otherwise, they are satisfied.

The bookstore owner knows a secret technique to remain not grumpy for <code>minutes</code> consecutive minutes, but this technique can only be used once.

Return the maximum number of customers that can be satisfied throughout the day.

&nbsp;
Example 1:

<div class="example-block">
Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], minutes = 3

Output: 16

Explanation:

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.
</div>

Example 2:

<div class="example-block">
Input: customers = [1], grumpy = [0], minutes = 1

Output: 1
</div>

&nbsp;
Constraints:

<ul>
	<li><code>n == customers.length == grumpy.length</code></li>
	<li><code>1 &lt;= minutes &lt;= n &lt;= 2 * 104</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>

maxSatisfied

Grumpy Bookstore Owner

Array

Sliding Window

有一个书店老板，他的书店开了&nbsp;<code>n</code>&nbsp;分钟。每分钟都有一些顾客进入这家商店。给定一个长度为 <code>n</code> 的整数数组 <code>customers</code> ，其中 <code>customers[i]</code> 是在第 <code>i</code> 分钟开始时进入商店的顾客数量，所有这些顾客在第 <code>i</code> 分钟结束后离开。

在某些分钟内，书店老板会生气。 如果书店老板在第 <code>i</code> 分钟生气，那么 <code>grumpy[i] = 1</code>，否则 <code>grumpy[i] = 0</code>。

当书店老板生气时，那一分钟的顾客就会不满意，若老板不生气则顾客是满意的。

书店老板知道一个秘密技巧，能抑制自己的情绪，可以让自己连续&nbsp;<code>minutes</code>&nbsp;分钟不生气，但却只能使用一次。

请你返回 这一天营业下来，最多有多少客户能够感到满意 。 
&nbsp;

示例 1：

<pre>
输入：customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], minutes = 3
输出：16
解释：书店老板在最后 3 分钟保持冷静。
感到满意的最大客户数量 = 1 + 1 + 1 + 1 + 7 + 5 = 16.
</pre>

示例 2：

<pre>
输入：customers = [1], grumpy = [0], minutes = 1
输出：1</pre>

&nbsp;

提示：

<ul>
	<li><code>n == customers.length == grumpy.length</code></li>
	<li><code>1 &lt;= minutes &lt;= n &lt;= 2 * 104</code></li>
	<li><code>0 &lt;= customers[i] &lt;= 1000</code></li>
	<li><code>grumpy[i] == 0 or 1</code></li>
</ul>