Grumpy Bookstore Owner

Companies:
  • Google Interview Questions
  • Nutanix Interview Questions

Grumpy Bookstore Owner

Companies: Nutanix, Google

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 :

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

Solution:

Using a sliding window approach will help us to solve this problem efficiently. Use a sliding window winOfMakeSatisfied to record the number of unsatisfied customers for X minutes. Deduct the unsatisfied customers from left end of the sliding window when it is wider than X, i.e. winOfMakeSatisfied -= grumpy[i - X] * customers[i - X]. Now, use satisfied to record the number of satistified customers without grumpy technique. After this iteration, our answer will be satisfied + max(winOfMakeSatisfied).

Implementation:

Java:


class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int X){
        int satisfied = 0, maxMakeSatisfied = 0;
        for (int i = 0, winOfMakeSatisfied = 0; i < grumpy.length; ++i) {
            if (grumpy[i] == 0) { satisfied += customers[i]; }
            else { winOfMakeSatisfied += customers[i]; }
            if (i >= X) {
                winOfMakeSatisfied -= grumpy[i - X] * customers[i - X];
            }
            maxMakeSatisfied = Math.max(winOfMakeSatisfied, maxMakeSatisfied);
        }
        return satisfied + maxMakeSatisfied;        
    }
}

Complexity Analysis:

  • Time Complexity: O(n)

  • Space Complexity: O(1)

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