# 1052.Grumpy-Bookstore-Owner

## 1052. Grumpy Bookstore Owner

## 题目地址

<https://leetcode.com/problems/grumpy-bookstore-owner/>

## 题目描述

```
Today, the bookstore owner has a store open for customers.length minutes.  Every minute, some number of customers (customers[i]) enter the store, and all those customers leave after the end of that minute.

On some minutes, the bookstore owner is grumpy.  If the bookstore owner is grumpy on the i-th minute, grumpy[i] = 1, otherwise grumpy[i] = 0.  When the bookstore owner is grumpy, the customers of that minute are not satisfied, otherwise they are satisfied.

The bookstore owner knows a secret technique to keep themselves not grumpy for X minutes straight, but can only use it once.

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

Example 1:
Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 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.

Note:
1 <= X <= customers.length == grumpy.length <= 20000
0 <= customers[i] <= 1000
0 <= grumpy[i] <= 1
```

## 代码

### Approach #1  Sliding Window

```java
class Solution {
  public int maxSatisfied(int[] customers, int[] grumpy, int X) {
        // 1. 计算原本就开心的客户数量 n base
    int baseCount = 0;
    for (int i = 0; i < customers.length; i++) {
      if (grumpy[i] == 0) {
        baseCount += customers[i];
      }
    }

    // 2. 生成 不开心客户的数组
    int[] sadCustomers = new int[customers.length];
    for (int i = 0; i < customers.length; i++) {
      if (grumpy[i] == 1) {
        sadCustomers[i] = customers[i];
      }
    }

    // 3. 计算 X 分钟板蓝根 所带来的最大收益
    // 3.1 先计算 前 X 分钟的收益
    int gain = 0;
    for (int i = 0; i < X; i++) {
      gain += sadCustomers[i];
    }

    // 3.2 采用 改变量 计算接下来每个区间的 收益并 记录最大值
    int maxGain = gain;
    for (int i = X; i < sadCustomers.length; i++) {
      gain += sadCustomers[i];
      gain -= sadCustomers[i - X];
      if (gain > maxGain) {
        maxGain = gain;
      }
    }

    // 4. 求和
    return baseCount + maxGain;
  }
}
```

### Approach #2

```java
class Solution {
  public int maxSatisfied(int[] customers, int[] grumpy, int X) {
    int maxGrumpy = 0;
    int happy = 0;
    int g = 0;
    for (int i = 0; i < customers.length; i++) {
      if (i >= X) {
        g = g - (customers[i - X] * grumpy[i - X]);
      }
      g += (customers[i] * grumpy[i]);
      maxGrumpy = Math.max(maxGrumpy, g);
      if (grumpy[i] == 0) {
        happy += customers[i];
      }
    }

    happy += maxGrumpy;
    return happy;
  }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/array/1052.grumpy-bookstore-owner.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.