LeetCode 1052: Grumpy Bookstore Owner Solution

Master LeetCode problem 1052 (Grumpy Bookstore Owner), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1052. Grumpy Bookstore Owner

Medium

Array Sliding Window

Problem Explanation

Explanation:

The problem requires us to find the maximum number of customers that can be satisfied throughout the day by using a secret technique to remain not grumpy for minutes consecutive minutes.
We can solve this problem by sliding window technique.
We will iterate through the customers array and calculate the total satisfied customers without using the secret technique.
Then, we will use a sliding window of size minutes to find the maximum number of additional customers that can be satisfied by using the secret technique.
The maximum satisfied customers will be the sum of total satisfied customers and the additional satisfied customers within the sliding window.

Time Complexity: O(n)
Space Complexity: O(1)

Solution Code

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        int n = customers.length;
        int satisfied = 0;
        int maxAdditionalSatisfied = 0;
        int additionalSatisfied = 0;
        
        for (int i = 0; i < n; i++) {
            if (grumpy[i] == 0) {
                satisfied += customers[i];
            }
        }
        
        for (int i = 0; i < minutes; i++) {
            if (grumpy[i] == 1) {
                additionalSatisfied += customers[i];
            }
        }
        
        maxAdditionalSatisfied = additionalSatisfied;
        
        for (int i = minutes; i < n; i++) {
            if (grumpy[i] == 1) {
                additionalSatisfied += customers[i];
            }
            if (grumpy[i - minutes] == 1) {
                additionalSatisfied -= customers[i - minutes];
            }
            maxAdditionalSatisfied = Math.max(maxAdditionalSatisfied, additionalSatisfied);
        }
        
        return satisfied + maxAdditionalSatisfied;
    }
}

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