Grumpy Bookstore OwnerLeetcode Problem 1052

英文原题

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.

中文翻译

今天，书店开店 customers.length 分钟。

每分钟，有 customers[i] 个人进入书店，并且在这分钟结束时全都离开了书店。

在有些时间（某几分钟），书店店长心情不好。

如果书店店长在 第 i 分钟心情不好，那么 grumpy[i] = 1，否则 grumpy[i] = 0。

当书店店长 心情不好的时候，那一分钟的客户也不开心，反之，客户就开心。

然而书店店长，知道服用板蓝根能让自己连续 X 分钟 不 心情不好（也就是一直心情好），但这板蓝根只能服用一次。

请你返回 这一天能让最多消费者开心的数量。

样例

Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 3
Output: 16
解释：店长决定在最后 3 分钟 服用 板蓝根
所以能让消费者开心的最多人数是 = 1 + 1 + 1 + 1 + 7 + 5 = 16.

问题签名

Java 代码

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class Solution {    public int maxSatisfied(int[] customers, int[] grumpy, int X) {     }}

Java Script 代码

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/*** @param {number[]} customers* @param {number[]} grumpy* @param {number} X* @return {number}*/var maxSatisfied = function(customers, grumpy, X) { };

Python 代码

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class Solution:    def maxSatisfied(self, customers: List[int], grumpy: List[int], X: int) -> int: