# leetcode\_1052

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

## Solutions

1. **maximum window**

```cpp
class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int X) {
        int t = customers.size(), maxg = 0, sumg = 0, res = 0;
        for (int i = 0; i < t; i++) {
            if (i - X >= 0)
                sumg -= grumpy[i - X] * customers[i - X];
            sumg += grumpy[i] * customers[i];
            if (sumg > maxg)
                maxg = sumg;
            res += (1 - grumpy[i]) * customers[i];
        }

        return res + maxg;
    }
};
```