Formatted question description: https://leetcode.ca/all/1052.html

# 1052. Grumpy Bookstore Owner (Medium)

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

Related Topics:
Array, Sliding Window

## Solution 1. Sliding Window

// OJ: https://leetcode.com/problems/grumpy-bookstore-owner/

// Time: O(N)
// Space: O(1)
class Solution {
public:
int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int X) {
int ans = 0, sum = 0;
for (int i = 0; i < customers.size(); ++i) sum += grumpy[i] == 0 ? customers[i] : 0;
for (int i = 0; i < customers.size(); ++i) {
if (grumpy[i] == 1) sum += customers[i];
if (i >= X && grumpy[i - X] == 1) sum -= customers[i - X];
if (i >= X - 1) ans = max(ans, sum);
}
return ans;
}
};


Java

class Solution {
public int maxSatisfied(int[] customers, int[] grumpy, int X) {
int satisfied = 0;
int length = customers.length;
for (int i = 0; i < length; i++) {
if (grumpy[i] == 0)
satisfied += customers[i];
}
int unsatisfiedWindow = 0;
for (int i = 0; i < X; i++) {
if (grumpy[i] == 1)
unsatisfiedWindow += customers[i];
}
int maxUnsatisfiedWindow = unsatisfiedWindow;
for (int i = X; i < length; i++) {
int prevIndex = i - X;
if (grumpy[prevIndex] == 1)
unsatisfiedWindow -= customers[prevIndex];
if (grumpy[i] == 1)
unsatisfiedWindow += customers[i];
maxUnsatisfiedWindow = Math.max(maxUnsatisfiedWindow, unsatisfiedWindow);
}
return satisfied + maxUnsatisfiedWindow;
}
}