1052. Grumpy Bookstore Owner

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1052. Grumpy Bookstore Owner

Description

There is a bookstore owner that has a store open for n minutes. You are given an integer array customers of length n where customers[i] is the number of the customers that enter the store at the start of the i^th minute and all those customers leave after the end of that minute.

During certain minutes, the bookstore owner is grumpy. You are given a binary array grumpy where grumpy[i] is 1 if the bookstore owner is grumpy during the i^th minute, and is 0 otherwise.

When the bookstore owner is grumpy, the customers entering during that minute are not satisfied . Otherwise, they are satisfied.

The bookstore owner knows a secret technique to remain not grumpy for minutes consecutive minutes, but this technique can only be used once .

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

Example 1:

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

Example 2:

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Input: customers = [1], grumpy = [0], minutes = 1

Output: 1

Constraints:

  • n == customers.length == grumpy.length
  • 1 <= minutes <= n <= 2 * 10^4
  • 0 <= customers[i] <= 1000
  • grumpy[i] is either 0 or 1.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int minutes) {
        int n = customers.size(), right = 0, max_s1 = 0;
        int sum[2]{};
        while (right < n) {
            sum[grumpy[right]] += customers[right];
            if (right >= minutes - 1) {
                max_s1 = max(max_s1, sum[1]);
                sum[1] -= grumpy[right - minutes + 1] ? customers[right - minutes + 1] : 0;
            }
            right++;
        }
        return sum[0] + max_s1;
    }
};