Number of Ways to Divide a Long Corridor - Practice Coding Problems

Number of Ways to Divide a Long Corridor - Problem

Library Corridor Division Challenge

Imagine you're an interior designer tasked with dividing a long library corridor into perfect reading sections! 📚

You're given a string corridor representing a line of seats ('S') and plants ('P'). Your goal is to install room dividers to create sections where each section contains exactly 2 seats and any number of plants.

🏗️ Setup: Room dividers are already installed at both ends (before index 0 and after index n-1). You can install additional dividers between any two adjacent positions.

🎯 Goal: Count how many different ways you can divide the corridor into valid sections. Two arrangements are different if there's at least one position where one has a divider and the other doesn't.

Return the count modulo 10⁹ + 7, or 0 if no valid division exists.

Input & Output

basic_example.py — Python

$ Input: corridor = "SSPPSPS"

› Output: 3

💡 Note: We have 4 seats total, so we need 2 sections with 2 seats each. The sections must be [SS][??] and [??][S]. The plant gaps between sections give us 3 ways to place the divider: after index 2, 3, or 4.

single_section.py — Python

$ Input: corridor = "PPSPSP"

› Output: 1

💡 Note: Only 2 seats total, so only one section needed. There's exactly 1 way to arrange this (no internal dividers needed).

impossible_case.py — Python

$ Input: corridor = "S"

› Output: 0

💡 Note: Only 1 seat, but each section needs exactly 2 seats. Impossible to create valid sections.

Visualization

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Understanding the Visualization

Count the Chairs

First, count total chairs. If odd number or zero, it's impossible to make pairs.

Find Chair Pairs

Identify where each 2-chair reading nook begins and ends.

Locate Plant Gardens

Between each pair of nooks, there's a garden area with plants.

Count Wall Options

In each plant garden, you can place the dividing wall at multiple positions.

Multiply Choices

Total arrangements = ways to place wall in garden 1 × ways in garden 2 × ...

Key Takeaway

🎯 Key Insight: Instead of trying every possible arrangement, count the placement options in each plant gap and multiply them together!

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to find seat positions and count plants

✓ Linear Growth

Space Complexity

O(1)

Only storing counters and positions

✓ Linear Space

Constraints

n == corridor.length
1 ≤ n ≤ 10⁵
corridor[i] is either 'S' or 'P'
Each section must contain exactly 2 seats

Asked in

G Google 24 a Amazon 18 f Meta 12 ⊞ Microsoft 8

The optimal solution uses mathematical counting instead of brute force. Key insight: find all seat positions, group them into pairs (sections), then count plant gaps between sections. Each gap provides multiple divider placement options - multiply these counts together for the final answer. Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Counting (Optimal)	O(n)	O(1)	Find seat pairs and count plant gaps between them

Mathematical Counting (Optimal) — Algorithm Steps

Count total seats - if not even or zero, return 0
Find positions of every 2nd seat (end of each section)
Count plants between consecutive section boundaries
Multiply the number of ways to place dividers in each gap

Visualization

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Step-by-Step Walkthrough

Locate Seat Pairs

Find positions where each 2-seat section must end

Identify Plant Gaps

Count plants between consecutive section boundaries

Count Placement Options

Each gap offers multiple divider placement positions

Multiply Results

Total ways = product of options in each gap

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int numberOfWays(char* corridor) {
    const int MOD = 1000000007;
    
    int seatPositions[1000];
    int seatCount = 0;
    
    for (int i = 0; i < strlen(corridor); i++) {
        if (corridor[i] == 'S') {
            seatPositions[seatCount++] = i;
        }
    }
    
    if (seatCount == 0 || seatCount % 2 != 0) {
        return 0;
    }
    
    if (seatCount == 2) {
        return 1;
    }
    
    long long result = 1;
    
    for (int i = 2; i < seatCount; i += 2) {
        int prevSectionEnd = seatPositions[i - 1];
        int currSectionStart = seatPositions[i];
        int gapSize = currSectionStart - prevSectionEnd;
        result = (result * gapSize) % MOD;
    }
    
    return (int) result;
}

int main() {
    char corridor[1000];
    scanf("\"%[^\"]\"", corridor);
    printf("%d\n", numberOfWays(corridor));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to find seat positions and count plants

✓ Linear Growth

Space Complexity

O(1)

Only storing counters and positions

✓ Linear Space

Constraints

n == corridor.length
1 ≤ n ≤ 10⁵
corridor[i] is either 'S' or 'P'
Each section must contain exactly 2 seats

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Mathematical Counting (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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