LeetCode-in-Java

1052. Grumpy Bookstore Owner

Medium

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

On some 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 ith minute, and is 0 otherwise.

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 minutes consecutive minutes, 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], 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:

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

Output: 1

Constraints:

• n == customers.length == grumpy.length
• 1 <= minutes <= n <= 2 * 104
• 0 <= customers[i] <= 1000
• grumpy[i] is either 0 or 1.

Solution

public class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        // storing numbers of customers who faced grumpy owner till ith minute.
        int[] grumpySum = new int[grumpy.length];
        int ans = 0;
        if (grumpy[0] == 1) {
            grumpySum[0] = customers[0];
        } else {
            ans += customers[0];
        }
        for (int i = 1; i < grumpy.length; i++) {
            if (grumpy[i] == 1) {
                grumpySum[i] = grumpySum[i - 1] + customers[i];
            } else {
                grumpySum[i] = grumpySum[i - 1];
                ans += customers[i];
            }
        }
        // calculating max number of customers who faced grumpy owner in a window of size 'minutes'.
        int max = 0;
        for (int i = 0; i <= customers.length - minutes; i++) {
            if (i == 0) {
                max = Math.max(max, grumpySum[i + minutes - 1]);
            } else {
                max = Math.max(max, grumpySum[i + minutes - 1] - grumpySum[i - 1]);
            }
        }
        // making the owner non-grumpy in that max window and adding the number of customers who do
        // not face the grumpy customers.
        ans += max;
        return ans;
    }
}

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