[leetcode] 1052. Grumpy Bookstore Owner

Description

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.

分析

题目的意思是：给你一个数组，把其中grumpy为0的位置加起来，然后给了你一个X，你可以把X范围内的数也加起来，求总的最大值。我第一次看题目没看懂，后面发现懂了，好像X就是一个窗口，除了为0的位置的数，还有滑动窗口里面的数也要加起来。我顿时看见这个慌了神，看了别人的参考代码才发现需要用一个滑动窗口找到X覆盖范围的最大值，然后重新加上grumpy位置为0的数就行了。

代码

class Solution:
def maxSatisfied(self, customers: List[int], grumpy: List[int], X: int) -> int:
i=0
max_save=0
cur=0
n=len(customers)
for j in range(n):
if(j-i+1>X):
if(grumpy[i]):
cur-=customers[i]
i+=1
if(grumpy[j]):
cur+=customers[j]
max_save=max(max_save,cur)

s=sum(customers[i] for i in range(n) if(grumpy[i]==0))
return s+max_save

参考文献

​​[LeetCode] Python 3 Sliding window​​