What is the time complexity of Number of Ways to Divide a Long Corridor?

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

What data structures are used to solve Number of Ways to Divide a Long Corridor?

This problem uses Math, String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Ways to Divide a Long Corridor?

Insight 1: The number of seats must be even for a valid division to exist. Insight 2: We only need to consider positions between pairs of seats where we can place a divider. Plants within a section don't affect the placement of dividers.

Which companies ask LeetCode 2147. Number of Ways to Divide a Long Corridor in interviews?

This problem is frequently asked at Zomato.

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Number of Ways to Divide a Long Corridor - LeetCode 2147 Solution

Q: How to solve Number of Ways to Divide a Long Corridor?

The solution iterates through the corridor string, counting the number of seats encountered. When we find the second seat of a section (seats_counted % 2 == 0), we then calculate the number of plant positions (or potential divider positions) until the next seat is found. The number of ways is the product of these plant position counts (plus one, because no divider can also be placed). If the total number of seats is odd or zero, there are no valid divisions, and we return 0. A modular arithmetic is used to prevent integer overflows.

Number of Ways to Divide a Long Corridor - Complete Solution Guide

Number of Ways to Divide a Long Corridor is LeetCode problem 2147, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Along a long library corridor, there is a line of seats and decorative plants. You are given a 0-indexed string corridor of length n consisting of letters 'S' and 'P' where each 'S' represents a seat and each 'P' represents a plant. One room divider has already been installed to the left of index 0 , and another to the right of index n - 1 . Additional room dividers can be installed. For each position between indices i - 1 and i ( 1 <= i <= n - 1 ), at most one divider can be installed. Divide t

Detailed Explanation

The problem asks us to divide a corridor represented by a string into sections, where each section contains exactly two seats ('S'). The corridor has dividers already installed at the beginning and end. We need to find the number of ways to place additional dividers between plants ('P') such that each section between dividers has exactly two seats. The final result should be modulo 10^9 + 7. If the number of seats is zero or odd, there's no valid division, and we should return 0.

Solution Approach

The solution iterates through the corridor string, counting the number of seats encountered. When we find the second seat of a section (seats_counted % 2 == 0), we then calculate the number of plant positions (or potential divider positions) until the next seat is found. The number of ways is the product of these plant position counts (plus one, because no divider can also be placed). If the total number of seats is odd or zero, there are no valid divisions, and we return 0. A modular arithmetic is used to prevent integer overflows.

Step-by-Step Algorithm

Step 1: Initialize `ans` to 1, `seats_counted` to 0, and `last_seat_idx` to -1.
Step 2: Iterate through the `corridor` string.
Step 3: If the current character is 'S', increment `seats_counted`.
Step 4: If `seats_counted` is greater than 2 and is odd, it means we've just encountered the first seat of a new pair after a completed section. Multiply `ans` by the number of positions between the current seat and the last seat (`i - last_seat_idx`). Take modulo `10**9 + 7`.
Step 5: Update `last_seat_idx` to the current index `i`.
Step 6: After iterating through the entire string, check if `seats_counted` is 0 or odd. If so, return 0.
Step 7: Return the value of `ans`.

Key Insights

Insight 1: The number of seats must be even for a valid division to exist.
Insight 2: We only need to consider positions between pairs of seats where we can place a divider. Plants within a section don't affect the placement of dividers.
Insight 3: The number of ways to divide the corridor is the product of the number of plant spaces between each pair of seat pairs.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, String, Dynamic Programming.

Companies

Asked at: Zomato.

Number of Ways to Divide a Long Corridor | LeetCode 2147

Hard

Math String Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int numberOfWays(String corridor) {
        int MOD = 1000000007;
        
        long ans = 1;
        int seats_counted = 0;
        int last_seat_idx = -1;

        for (int i = 0; i < corridor.length(); i++) {
            char charAtI = corridor.charAt(i);
            if (charAtI == 'S') {
                seats_counted += 1;
                
                if (seats_counted > 2 && seats_counted % 2 == 1) {
                    ans = (ans * (i - last_seat_idx)) % MOD;
                }
                
                last_seat_idx = i;
            }
        }
        
        if (seats_counted == 0 || seats_counted % 2 != 0) {
            return 0;
        }
        
        return (int) ans;
    }
}

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Problem Info

Description

Problem 2147: Number of Ways to Divide a Long Corridor - Detailed explanation and solution approaches.

Topics

Math String Dynamic Programming

Companies

Zomato

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2147

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

7 steps

1
Step 1: Initialize `ans` to 1, `seats_counted` to 0, and `last_seat_idx` to -1.
2
Step 2: Iterate through the `corridor` string.
3
Step 3: If the current character is 'S', increment `seats_counted`.
4
Step 4: If `seats_counted` is greater than 2 and is odd, it means we've just encountered the first seat of a new pair after a completed section. Multiply `ans` by the number of positions between the current seat and the last seat (`i - last_seat_idx`). Take modulo `10**9 + 7`.
5
Step 5: Update `last_seat_idx` to the current index `i`.
6
Step 6: After iterating through the entire string, check if `seats_counted` is 0 or odd. If so, return 0.
7
Step 7: Return the value of `ans`.

💡

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 iterate over the string once. The space complexity is constant because the amount of memory used does not scale 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

Dynamic Programming

While a DP approach is possible, it's generally less efficient than the straightforward iterative approach used in the provided solution. A DP solution could involve tracking the number of ways to divide the corridor up to a certain index, considering whether the current character is a seat or a plant.

Time: O(n), where n is the length of the corridor string.Space: O(n), to store the DP table.

Trade-offs: The iterative approach is preferred because it's simpler to implement and has a lower space complexity (O(1) vs O(n)). The DP approach might be useful if the problem constraints were different or if we wanted to calculate additional information about the divisions.

✨

Key Insights

1Insight 1: The number of seats must be even for a valid division to exist.
2Insight 2: We only need to consider positions between pairs of seats where we can place a divider. Plants within a section don't affect the placement of dividers.
3Insight 3: The number of ways to divide the corridor is the product of the number of plant spaces between each pair of seat pairs.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: `corridor` = "S". The code correctly returns 0 because there are not enough seats to form a valid section.
Edge case 2: `corridor` = "P". The code correctly returns 0 because there are no seats.
Edge case 3: `corridor` = "SPPPSP". The code correctly returns 0 because the total number of seats is odd (only one).
Edge case 4: `corridor` = "SSPPSPS". The code correctly computes the number of ways (3).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the number of seats is zero or odd. This leads to incorrect results or runtime errors.

✗Common mistake 2: Integer overflow when calculating the number of ways. Remembering to use modular arithmetic with a modulo of 10^9 + 7 is crucial.

✗Common mistake 3: Incorrectly calculating the number of possible divider placements between seat pairs. It's the difference in indices between the next seat and the last seat of the previous pair.

✗Common mistake 4: Not initializing the answer (`ans`) to 1. Since we are multiplying, starting with 0 will always result in 0.

✅

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: Resource allocation: Imagine you are allocating resources in a production line. Each seat ('S') represents a critical station, and plants ('P') are buffer areas. You want to divide the line into segments with two critical stations each, with flexibility in dividing the buffer zones.

🏦Real-world scenario 2: Network segmentation: Think of 'S' as servers requiring isolation and 'P' as network devices. The algorithm can help determine the number of ways to segment the network such that each isolated segment contains two critical servers.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: November 15, 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 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Number of Ways to Divide a Long Corridor"?

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

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