What is the time complexity of Minimum Penalty for a Shop?

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

What data structures are used to solve Minimum Penalty for a Shop?

This problem uses String, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Penalty for a Shop?

Insight 1: Instead of calculating the penalty for each possible closing time independently, we can use a running score (prefix sum technique) to efficiently update the penalty as we iterate through the hours. Insight 2: The problem is essentially finding the prefix where we maximize the difference between the number of 'Y's encountered before the prefix and the number of 'N's encountered after the prefix (or vice versa minimizing the penalty).

Which companies ask LeetCode 2483. Minimum Penalty for a Shop in interviews?

This problem is frequently asked at Stripe.

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Minimum Penalty for a Shop - LeetCode 2483 Solution

Q: How to solve Minimum Penalty for a Shop?

The provided solution uses a single pass through the 'customers' string to calculate a 'current_score'. For each 'Y', the score is incremented, and for each 'N', it is decremented. The 'max_score' keeps track of the highest score encountered so far, and 'best_hour' is updated whenever a new 'max_score' is found. Initially, the max_score and best_hour are set to 0, representing closing at the beginning. The algorithm efficiently finds the hour that maximizes the current_score, effectively minimizing the penalty.

Minimum Penalty for a Shop - Complete Solution Guide

Minimum Penalty for a Shop is LeetCode problem 2483, 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

You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters 'N' and 'Y' : if the i th character is 'Y' , it means that customers come at the i th hour whereas 'N' indicates that no customers come at the i th hour. If the shop closes at the j th hour ( 0 <= j <= n ), the penalty is calculated as follows: For every hour when the shop is open and no customers come, the penalty increases by 1 . For every hour when the shop is closed and cu

Detailed Explanation

The problem asks us to find the optimal closing time for a shop to minimize the penalty based on customer visits. We're given a string 'customers' where 'Y' indicates customers visited during that hour and 'N' indicates no customers visited. The penalty is calculated as follows: for every hour the shop is open when no customers come, the penalty increases by 1, and for every hour the shop is closed when customers come, the penalty increases by 1. The goal is to return the earliest hour at which the shop should be closed to incur the minimum penalty.

Solution Approach

The provided solution uses a single pass through the 'customers' string to calculate a 'current_score'. For each 'Y', the score is incremented, and for each 'N', it is decremented. The 'max_score' keeps track of the highest score encountered so far, and 'best_hour' is updated whenever a new 'max_score' is found. Initially, the max_score and best_hour are set to 0, representing closing at the beginning. The algorithm efficiently finds the hour that maximizes the current_score, effectively minimizing the penalty.

Step-by-Step Algorithm

Step 1: Initialize max_score, best_hour, and current_score to 0.
Step 2: Iterate through the 'customers' string from left to right.
Step 3: For each character, if it's 'Y', increment current_score; otherwise, decrement current_score.
Step 4: If the current_score is greater than the max_score, update max_score to current_score and best_hour to the current hour index plus 1.
Step 5: After iterating through the entire string, return best_hour.

Key Insights

Insight 1: Instead of calculating the penalty for each possible closing time independently, we can use a running score (prefix sum technique) to efficiently update the penalty as we iterate through the hours.
Insight 2: The problem is essentially finding the prefix where we maximize the difference between the number of 'Y's encountered before the prefix and the number of 'N's encountered after the prefix (or vice versa minimizing the penalty).
Insight 3: Closing at hour 0 means the shop is immediately closed, and closing at the last hour means the shop stays open throughout the entire period.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Prefix Sum.

Companies

Asked at: Stripe.

Minimum Penalty for a Shop | LeetCode 2483

Medium

String Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int bestClosingTime(String customers) {
        int max_score = 0;
        int best_hour = 0;
        int current_score = 0;
        
        for (int i = 0; i < customers.length(); i++) {
            char status = customers.charAt(i);
            if (status == 'Y') {
                current_score += 1;
            } else {
                current_score -= 1;
            }
            
            if (current_score > max_score) {
                max_score = current_score;
                best_hour = i + 1;
            }
        }
        
        return best_hour;
    }
}

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Problem Info

Description

Problem 2483: Minimum Penalty for a Shop - Detailed explanation and solution approaches.

Topics

String Prefix Sum

Companies

Stripe

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2483

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

5 steps

1
Step 1: Initialize max_score, best_hour, and current_score to 0.
2
Step 2: Iterate through the 'customers' string from left to right.
3
Step 3: For each character, if it's 'Y', increment current_score; otherwise, decrement current_score.
4
Step 4: If the current_score is greater than the max_score, update max_score to current_score and best_hour to the current hour index plus 1.
5
Step 5: After iterating through the entire string, return best_hour.

💡

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 linear because we process each character in the input string once. The space complexity is constant because the amount of memory used doesn't depend on 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 with two arrays

Create two prefix sum arrays: one for the number of 'Y's and one for the number of 'N's. Then, iterate through all possible closing times and calculate the penalty based on the prefix sums. This would entail one array holding number of 'Y' before some index i, and another array holding the number of 'N' after the index i. Calculate the penalty for each closing time and save the index with the minimum penalty

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

Trade-offs: This approach is more readable for some, but requires O(n) extra space for the prefix sum arrays. The single-pass approach is more space-efficient.

✨

Key Insights

1Insight 1: Instead of calculating the penalty for each possible closing time independently, we can use a running score (prefix sum technique) to efficiently update the penalty as we iterate through the hours.
2Insight 2: The problem is essentially finding the prefix where we maximize the difference between the number of 'Y's encountered before the prefix and the number of 'N's encountered after the prefix (or vice versa minimizing the penalty).
3Insight 3: Closing at hour 0 means the shop is immediately closed, and closing at the last hour means the shop stays open throughout the entire period.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: 'customers' is all 'Y's. The solution correctly returns the length of the string as the best closing time, because it is better to stay open and serve all customers than closing and receiving penalty.
Edge case 2: 'customers' is all 'N's. The solution correctly returns 0. It is better to close the shop immediately if no customer arrives.
Edge case 3: 'customers' is empty string. This is not a valid input according to the constraints (1 <= customers.length <= 10^5). But if it was, the algorithm needs to handle it gracefully, potentially returning 0 or raising an exception as appropriate based on the specific problem definition.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the penalty. Forgetting that the penalty increases when the shop is open but no customers come or closed but customers do come.

✗Common mistake 2: Not handling the edge cases correctly, especially when the input contains only 'Y's or only 'N's. Incorrectly assuming that closing at a particular index means the shop is open up to and including that index.

✅

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 times in call centers or help desks. Determining the optimal time to close a service based on historical demand data.

🏦Real-world scenario 2: Deciding when to shut down a production line for maintenance. Balancing the cost of downtime with the need for maintenance and expected production demand.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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 22, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Penalty for a Shop"?

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

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