How do you solve Grumpy Bookstore Owner on LeetCode?

Grumpy Bookstore Owner (LeetCode #1052) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using a sliding window technique allows us to efficiently calculate the maximum additional customers satisfied without re-evaluating the entire array.

What is the brute force solution for Grumpy Bookstore Owner?

The brute force approach for Grumpy Bookstore Owner is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will try every possible starting minute for the bookstore owner to use their non-grumpy technique and calculate the total satisfied customers for each case. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Grumpy Bookstore Owner?

Grumpy Bookstore Owner uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Sliding Window, Array.

What are the interview tips for Grumpy Bookstore Owner?

Always clarify the constraints and edge cases with your interviewer. Think aloud while solving the problem to show your thought process. Practice explaining your approach clearly and concisely.

What are common mistakes when solving Grumpy Bookstore Owner?

Failing to account for the case when the non-grumpy period overlaps with the end of the array. Not correctly summing the satisfied customers initially before applying the sliding window.

#1052

Grumpy Bookstore Owner

Medium

ArraySliding WindowSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

We can use a sliding window approach to efficiently calculate the maximum number of customers that can be satisfied by using the non-grumpy technique for a fixed number of minutes. This reduces the need for nested loops.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total number of satisfied customers without using the non-grumpy technique.
2Step 2: Use a sliding window of size 'minutes' to find the maximum number of customers that can be added to the satisfied count by considering the grumpy minutes.
3Step 3: Slide the window across the array, updating the total satisfied customers accordingly.

solution.py14 lines

1# Full working Python code
2
3def maxSatisfied(customers, grumpy, minutes):
4    total_satisfied = sum(customers[i] for i in range(len(customers)) if grumpy[i] == 0)
5    max_additional = 0
6    current_additional = 0
7    for i in range(len(customers)):
8        if grumpy[i] == 1:
9            current_additional += customers[i]
10        if i >= minutes:
11            if grumpy[i - minutes] == 1:
12                current_additional -= customers[i - minutes]
13        max_additional = max(max_additional, current_additional)
14    return total_satisfied + max_additional

ℹ

Complexity note: This complexity is linear because we traverse the customers array only a couple of times, making it much more efficient than the brute-force approach.

1Using a sliding window technique allows us to efficiently calculate the maximum additional customers satisfied without re-evaluating the entire array.
2Understanding the problem in terms of satisfied vs. unsatisfied customers helps in formulating the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.