Number of Ways to Divide a Long Corridor

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;
    }
};

The provided C++ solution is designed to solve the problem of counting the number of valid path divisions in a pathway, where sections are divided by 'S' characters representing seats. The corridor is represented as a string where each character can either be a 'S' or 'P', with 'S' standing for seats and 'P' for plants. The objective is to determine the number of ways to divide the corridor into sections, each containing exactly two 'S'.

Key aspects of the provided code include:

• The countPaths function initiates by preparing essential variables:

  • MOD = 1000000007 helps in keeping results within integer limits using modulo operations to avoid overflow.
  • ways tracks the total number of ways to divide the corridor up to the current section of processing and starts with an initial value of 1 to represent a valid single way.
  • sectionSeats counts the seats encountered and resets after every two seats.
  • lastSIndex keeps track of the index of the second seat in the last counted section.

• The logic iterates through each character in the string:

  • When a seat ('S') is encountered, it increases the sectionSeats counter.
  • If sectionSeats hits 2 (indicating the end of a valid section), it updates lastSIndex with the current index and resets sectionSeats to begin counting for a new section.
  • If it's the first seat encountered after a previously completed section (sectionSeats == 1 and lastSIndex != -1), it computes the number of ways considering the seats' positions since lastSIndex.

• The multiplier ways captures the number of variations possible from one seating position to the next by considering the positions of intermediate plant characters as potential starting points for new sections.

• The function concludes by verifying the conditions:

  • It returns 0 if an incomplete section remains (i.e., a single 'S' that hasn't formed a pair).
  • It also returns 0 if no complete sections were found.

• Finally, it returns the total count of division ways considering all possible placements, calculated under modulo operations to handle large numbers.

This solution is efficient in terms of both time and space complexity, iterating through the string in linear time (O(n)), while only using a constant amount of extra space (O(1)). It gracefully handles edge cases like strings without 'S' or strings with an uncomfortable number of 'S' to avoid unexpected sections. Overall, this implementation provides a robust and optimized approach to solving the problem using modular arithmetic and efficient counting mechanisms.

class Solution {
    public int countWays(String path) {
        // Modulus value for avoiding large number overflow
        final int MODULO = 1_000_000_007;

        // Variable to store the result of possible arrangements
        long ways = 1;

        // Count of seats found in the current segment
        int seatCounter = 0;

        // Save last seen seat's index
        Integer lastSectionEnd = null;

        // Processing each character in the corridor
        for (int i = 0; i < path.length(); i++) {
            if (path.charAt(i) == 'S') {
                seatCounter++;

                // Handle the scenario when a pair is completed
                if (seatCounter == 2) {
                    lastSectionEnd = i;
                    seatCounter = 0; // Reset for new segment
                }
                // Only process multiplication if at least one pair is found
                else if (seatCounter == 1 && lastSectionEnd != null) {
                    ways *= (i - lastSectionEnd);
                    ways %= MODULO; // Keep the result within bounds
                }
            }
        }

        // Conditions if impossible to appropriately partition
        if (seatCounter == 1 || lastSectionEnd == null) {
            return 0;
        }

        // Final number of arrangements
        return (int) ways;
    }
}

This Java solution aims to solve the problem of counting the possible ways to segment a long corridor based on seats ('S') and plants ('P'). The method countWays receives a String path representing the layout of the corridor.

Here's how the solution operates:

• Declare and initialize a modulus constant MODULO to prevent overflow during calculations due to large numbers.
• Use a long variable, ways, initialized to 1, to hold the result.
• Use int seatCounter to keep track of the number of seats found in the current segment.
• Use Integer lastSectionEnd to store the index of the last seat in the previous successfully counted segment.

The code processes each character in the path:

• Increments seatCounter when a seat ('S') is encountered.
• Every two seats detected (when seatCounter equals 2), update lastSectionEnd to the current index and reset seatCounter to 0 for the next potential segment.
• Multiplies ways by the distance between current and last found pairs of seats if a single seat is found following an earlier pair, updating the result modulo MODULO.

Special cases are handled:

• Return 0 if ending with an unpaired seat or if no seats are found throughout the corridor.

At the end of the loop, if the corridor is appropriately segmented, the method returns the number of possible arrangements cast to an int, ensuring it fits expected variable sizes. Thus, the solution efficiently partitions the corridor and computes all segmentations without direct iteration over vast sections repeatedly, making it computationally efficient.

int calculatePaths(char* path){
    const int MOD_CONSTANT = 1000000007;
    long ways = 1;
    int seatCount = 0;
    int lastValidIndex = -1;
    int pathLength = strlen(path);

    for (int i = 0; i < pathLength; i++) {
        if (path[i] == 'S') {
            seatCount++;

            if (seatCount == 2) {
                lastValidIndex = i;
                seatCount = 0;
            } else if (seatCount == 1 && lastValidIndex != -1) {
                ways *= (i - lastValidIndex);
                ways %= MOD_CONSTANT;
            }
        }
    }

    if (seatCount == 1 || lastValidIndex == -1) {
        return 0;
    }

    return (int) ways;
}

In the provided C code, you calculate the number of ways to divide a long corridor into sections based on the placement of seats represented by 'S' in a given path string. The method utilizes a modulo constant to prevent overflow issues during computation. Here's how the code accomplishes the task:

• Initialize variables to count the number of ways (ways), count seats (seatCount), and store the index of the last valid seat position (lastValidIndex).
• Iterate over the characters in the path. For each seat ('S') encountered:
  • Increment the seatCount.
  • If two seats are counted (seatCount reaches 2), reset the seatCount to zero and update lastValidIndex to the current index.
  • If one seat is counted and it's not the first seat in the path (lastValidIndex is not -1), calculate the product of the number of ways and the distance between the current and the last valid seat index. Apply the modulo operation to keep the number manageable.
• After iterating through the path, if the seatCount does not return to zero after the last seat or no valid seats were found (lastValidIndex is -1), return 0 as no valid division can be made.
• Otherwise, return the total number of ways calculated as an integer.

The crux of the solution lies in efficiently counting possible divisions using seat placements while adhering to modular arithmetic to handle large numbers. The approach ensures that the solution remains computationally feasible and precise for large input strings.

var countWays = function(path) {
    const MODULO = (10 ** 9) + 7;

    let ways = 1;
    let seatCount = 0;
    let lastSIndex = -1;

    for (let i = 0; i < path.length; i++) {
        if (path[i] == 'S') {
            seatCount++;

            if (seatCount == 2) {
                lastSIndex = i;
                seatCount = 0;
            } else if (seatCount == 1 && lastSIndex != -1) {
                ways *= (i - lastSIndex);
                ways %= MODULO;
            }
        }
    }

    if (seatCount == 1 || lastSIndex == -1) {
        return 0;
    }

    return ways;
};

In the solution for the problem of finding the number of ways to divide a long corridor, the JavaScript function countWays is utilized to compute the possible configurations.

• Initialize constants and variables:

  • A modulus MODULO is defined for handling large numbers, set to (10^9 + 7).
  • ways is initialized to 1, representing the number of ways to partition the corridor.
  • seatCount keeps track of the consecutive seats encountered.
  • lastSIndex stores the index of the last seat in a valid section.

• Process every character in the path string:

  • When encountering a seat ('S'):
    • Increment the seatCount.
    • If two seats have been counted:
      • Update lastSIndex to the current index and reset seatCount for the next potential section.
    • If only one seat has been counted and there is a preceding valid section (lastSIndex is not -1):
      • Calculate possible ways of dividing between the current seat and the last seat index, updating the ways variable and applying the modulus for large number handling.

• Final evaluation:

  • If after processing all characters only one seat remains unchecked or no valid section has been established (lastSIndex is -1), return 0 as no valid division is possible.
  • Otherwise, return the value of ways, representing the total number of valid partitioning configurations.

This solution efficiently considers every possible partitioning between valid sections of seats while ensuring operations remain within computational limits using modulus arithmetic.

class Solution:
    def waysToPartition(self, section: str) -> int:
        # Defining the modulus for large numbers
        modulus = 10**9 + 7

        # Initialize result multipler
        ways = 1

        # Temp counter for 'S'
        temp_seats = 0

        # Last index recorded for 'S'
        last_seat_index = None

        # Loop through corridor elements
        for idx, place in enumerate(section):
            if place == "S":
                temp_seats += 1

                # Handle complete pair of 'S'
                if temp_seats == 2:
                    last_seat_index = idx
                    temp_seats = 0
                
                # Handling pairs calculation
                elif temp_seats == 1 and last_seat_index is not None:
                    ways *= (idx - last_seat_index)
                    ways %= modulus
        
        # Check for unmatch pair or no pairs
        if temp_seats == 1 or last_seat_index is None:
            return 0

        # Return calculated ways
        return ways

In the Python3 solution provided, the goal is to determine the number of ways a long corridor can be partitioned based on specific conditions involving seat pairs marked by 'S'. The implementation includes the following key steps:

1. Initialize a modulus to handle large number calculations, set to (10^9 + 7).
2. Initialize ways to 1, which is used to store the total number of partitioning ways.
3. Utilize temp_seats to count temporary occurrences of 'S' and last_seat_index to store the last index where 'S' was encountered.
4. Iterate through the given corridor sections using a loop.
5. For each seat 'S' found, increment the temp_seats counter.
6. If temp_seats reaches 2, indicating a pair, update last_seat_index and reset temp_seats.
7. If temp_seats is 1 (indicating the start of a potential new pair) and last_seat_index is not None, calculate the number of ways by multiplying the current number of ways with the distance from the last_seat_index to the current index, then apply the modulus to handle large values.
8. At the end of the iteration, if there is an unmatched 'S' left or if no 'S' was found throughout the section (last_seat_index is None), the function returns 0 suggesting no possible partitions.

The function finally returns ways, indicating the total number of ways to partition the corridor satisfying the conditions of pairs of seats. This solution handles edge cases like unmatched seats and sections completely devoid of 'S', ensuring accurate and efficient computation for possible partitions.