Number of Ways to Divide a Long Corridor

Problem Statement

In this problem, you are tasked to partition a long library corridor lined with seats (denoted by 'S') and decorative plants (denoted by 'P'). The corridor is represented as a string corridor where 'S' and 'P' represent seats and plants respectively. Your aim is to install dividers such that you can segment the corridor into multiple non-overlapping sections. Each section must contain exactly two seats and could contain any number of plants. The configuration of this segmentation can lead to multiple valid arrangements.

Initially, dividers are placed at the very start (before index 0) and at the very end (after the last index). Additional dividers can be placed between any two characters in the string. You need to calculate the number of distinct ways to divide the corridor under the given rules. Due to potential large results, the output should be the count modulo 10^9 + 7. If it's not possible to segment the corridor as required, the function should return 0.

The challenge not only involves generating all possible valid partitions but also understanding how placements of dividers can differ yet result in the same segmentations, or differ to create unique valid segmentations.

Examples

Example 1

Input:

corridor = "SSPPSPS"

Output:

3

Explanation:

There are 3 different ways to divide the corridor.
The black bars in the above image indicate the two room dividers already installed.

Note that in each of the ways, each section has exactly two seats.

Example 2

Input:

corridor = "PPSPSP"

Output:

1

Explanation:

There is only 1 way to divide the corridor, by not installing any additional dividers.
Installing any would create some section that does not have exactly two seats.

Example 3

Input:

corridor = "S"

Output:

0

Explanation:

There is no way to divide the corridor because there will always be a section that does not have exactly two seats.

Constraints

• n == corridor.length
• 1 <= n <= 105
• corridor[i] is either 'S' or 'P'.

Approach and Intuition

1. Counting Seats and Determining Feasibility:

   • Start by counting the total number of seats ('S'). If the count is odd or less than two, immediately return 0 since it's not possible to form sections with exactly two seats each.

2. Placement of Dividers After Every Two Seats:

   • Traverse the corridor string while keeping track of the number of seats encountered. Every time you count two seats, consider the possibility of placing a divider.
   • Count ways to place dividers between sections created every time two seats are grouped. This placement must consider all plants ('P') that can be included in these sections.

3. Multiplicative Combinations for Divider Placements:

   • Inserting a divider is optional but must adhere to the rule of exactly two seats per section. For each possible point to place a divider (right after two seats are counted), calculate the number of ways forward without disturbing the structure of two seats per section already counted.

4. Dynamic Programming for Count Management:

   • Use dynamic programming to remember the number of ways the corridor can be divided up to a certain point. This array updates as you process different valid divisions, thus building up on previous calculations.

5. Final Calculation and Modulo Operation:

   • As you iterate to the end of the string, maintain all possibilities valid up to that point under modulo 10^9 + 7 to manage large numbers and prevent overflow.

This approach allows for constructing segments progressively while counting the number of possible configurations dynamically, leading to an efficient solution even for longer strings, respecting the constraints.

Solutions

class Solution {
public:
    int countPaths(string path) {
        const int MOD = 1000000007;
        long ways = 1;
        int sectionSeats = 0;
        int lastSIndex = -1;
    
        for (int i = 0; i < path.length(); i++) {
            if (path[i] == 'S') {
                sectionSeats++;
    
                if (sectionSeats == 2) {
                    lastSIndex = i;
                    sectionSeats = 0;
                } else if (sectionSeats == 1 && lastSIndex != -1) {
                    ways *= (i - lastSIndex);
                    ways %= MOD;
                }
            }
        }
    
        if (sectionSeats == 1 || lastSIndex == -1) return 0;
    
        return (int) ways;
    }
};