What is the time complexity of Grumpy Bookstore Owner?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Grumpy Bookstore Owner?

This problem uses Array, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Grumpy Bookstore Owner?

Insight 1: The problem can be broken down into two parts: calculating the initially satisfied customers (when `grumpy[i] == 0`) and finding the maximum additional customers we can satisfy by using the technique. Insight 2: A sliding window approach is suitable for finding the maximum additional customers because we need to consider consecutive minutes for the technique.

Which companies ask LeetCode 1052. Grumpy Bookstore Owner in interviews?

This problem is frequently asked at Nutanix.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Grumpy Bookstore Owner - LeetCode 1052 Solution

Grumpy Bookstore Owner - Complete Solution Guide

Grumpy Bookstore Owner is LeetCode problem 1052, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 gru

Detailed Explanation

The problem describes a bookstore owner who is sometimes grumpy. We're given two arrays: `customers` representing the number of customers entering the store each minute, and `grumpy` indicating whether the owner is grumpy (1) or not (0) during that minute. A grumpy owner makes customers unsatisfied. We are allowed to use a special technique to keep the owner non-grumpy for `minutes` consecutive minutes, but only once. The goal is to find the maximum number of satisfied customers possible.

Solution Approach

The solution uses a sliding window technique. First, it calculates the number of customers who are already satisfied when the owner is not grumpy. Then, it uses a sliding window of size `minutes` to iterate through the `customers` array. The sliding window tracks the additional number of customers that can be satisfied by applying the technique during that specific window of `minutes`. We track the maximum additional customers possible. Finally, the maximum possible additional customers are added to the initially satisfied customers, and this sum is the final answer.

Step-by-Step Algorithm

Step 1: Calculate the number of initially satisfied customers by iterating through the `customers` and `grumpy` arrays. Add `customers[i]` to `initially_satisfied` if `grumpy[i]` is 0.
Step 2: Initialize `current_gain` to the number of customers that can be satisfied by using the technique for the first `minutes` minutes. Iterate through the first `minutes` indices, adding `customers[i]` to `current_gain` if `grumpy[i]` is 1.
Step 3: Initialize `max_gain` with `current_gain`.
Step 4: Iterate through the remaining minutes using a sliding window. For each minute `i` from `minutes` to `n-1`: Update `current_gain` by adding `customers[i]` if `grumpy[i]` is 1 and subtracting `customers[i - minutes]` if `grumpy[i - minutes]` is 1. This maintains the current 'gain' of using the technique for the current window.
Step 5: Update `max_gain` with the maximum of `max_gain` and `current_gain`.
Step 6: Return the sum of `initially_satisfied` and `max_gain`.

Key Insights

Insight 1: The problem can be broken down into two parts: calculating the initially satisfied customers (when `grumpy[i] == 0`) and finding the maximum additional customers we can satisfy by using the technique.
Insight 2: A sliding window approach is suitable for finding the maximum additional customers because we need to consider consecutive minutes for the technique.
Insight 3: The key is to separate the already satisfied customers from the potential gain by using the not-grumpy technique. Then, combine both at the end to calculate the maximum satisfaction.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Sliding Window.

Companies

Asked at: Nutanix.

Grumpy Bookstore Owner | LeetCode 1052

Medium

Array Sliding Window

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

Solution Code

Available in:

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        int n = customers.length;

        int initially_satisfied = 0;
        for (int i = 0; i < n; i++) {
            if (grumpy[i] == 0) {
                initially_satisfied += customers[i];
            }
        }

        int current_gain = 0;
        for (int i = 0; i < minutes; i++) {
            if (grumpy[i] == 1) {
                current_gain += customers[i];
            }
        }

        int max_gain = current_gain;

        for (int i = minutes; i < n; i++) {
            if (grumpy[i] == 1) {
                current_gain += customers[i];
            }
            if (grumpy[i - minutes] == 1) {
                current_gain -= customers[i - minutes];
            }

            max_gain = Math.max(max_gain, current_gain);
        }

        return initially_satisfied + max_gain;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1052: Grumpy Bookstore Owner - Detailed explanation and solution approaches.

Topics

Array Sliding Window

Companies

Nutanix

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1052

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Calculate the number of initially satisfied customers by iterating through the `customers` and `grumpy` arrays. Add `customers[i]` to `initially_satisfied` if `grumpy[i]` is 0.
2
Step 2: Initialize `current_gain` to the number of customers that can be satisfied by using the technique for the first `minutes` minutes. Iterate through the first `minutes` indices, adding `customers[i]` to `current_gain` if `grumpy[i]` is 1.
3
Step 3: Initialize `max_gain` with `current_gain`.
4
Step 4: Iterate through the remaining minutes using a sliding window. For each minute `i` from `minutes` to `n-1`: Update `current_gain` by adding `customers[i]` if `grumpy[i]` is 1 and subtracting `customers[i - minutes]` if `grumpy[i - minutes]` is 1. This maintains the current 'gain' of using the technique for the current window.
5
Step 5: Update `max_gain` with the maximum of `max_gain` and `current_gain`.
6
Step 6: Return the sum of `initially_satisfied` and `max_gain`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because we iterate through the input arrays of size 'n' a fixed number of times (specifically, calculating initially_satisfied, initializing the first window of size 'minutes', and sliding the window across the rest of the input). These are all linear operations. The space complexity is O(1) because we use only a few integer variables to store intermediate results. The number of variables doesn't grow with the size of the input.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Prefix Sum Array

Calculate a prefix sum array for the 'grumpy' customers. Then iterate through the array, calculating the difference between prefix sums to find the maximum gain for a window of 'minutes'.

Time: O(n)Space: O(n)

Trade-offs: Uses O(n) space for the prefix sum array, but can be more readable to some. The sliding window approach generally is preferable due to O(1) space complexity.

✨

Key Insights

1Insight 1: The problem can be broken down into two parts: calculating the initially satisfied customers (when `grumpy[i] == 0`) and finding the maximum additional customers we can satisfy by using the technique.
2Insight 2: A sliding window approach is suitable for finding the maximum additional customers because we need to consider consecutive minutes for the technique.
3Insight 3: The key is to separate the already satisfied customers from the potential gain by using the not-grumpy technique. Then, combine both at the end to calculate the maximum satisfaction.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: `minutes` is equal to or greater than `n`. In this case, all grumpy minutes can be converted to non-grumpy. The `max_gain` calculation correctly handles this case by covering the entire range. The loop `for (int i = minutes; i < n; i++)` will not execute because the loop condition won't be met. `max_gain` will be initialized with the gain from the first 'minutes' (or 'n' if 'minutes' > 'n') elements.
Edge case 2: All entries in `grumpy` are 0. In this case, `max_gain` will be 0 and the result will be `initially_satisfied`, which is the sum of all customers. The logic functions correctly without modification.
Edge case 3: All entries in `grumpy` are 1. The `initially_satisfied` will be 0, and the max_gain is calculated and returned, corresponding to the best contiguous 'minutes' span of the customer array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not accounting for the initially satisfied customers. Forgetting to add these to the maximum gain from using the technique.

✗Common mistake 2: Incorrectly calculating the gain within the sliding window. Making sure the window is of size `minutes` and correctly adding and subtracting elements is essential.

✗Common mistake 3: Not handling the case when minutes >= n correctly.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing service in a shop. Imagine a delivery service where the delivery time might be unreliable. Knowing the demand for deliveries and the 'unreliable' period, you can optimize a truck's route to maximize delivered packages during this period.

🏦Real-world scenario 2: Resource allocation during maintenance periods. A system administrator needs to schedule a maintenance period. They can use this approach to find the optimal time slot (of length 'minutes') that minimizes the impact on user activity, where 'customers' represents the user activity and 'grumpy' represents when maintenance is disruptive.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I determine the window size in sliding window problems?

For fixed-size windows (like max sum of k elements), window size is given. For variable-size windows (like smallest subarray with sum >= target), expand right pointer to meet condition, then shrink left pointer to find optimal. Track the best result during shrinking.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Grumpy Bookstore Owner"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether sliding window techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?