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 the corridor into non-overlapping sections, where each section has exactly two seats with any number of plants. There may be multiple ways to perform the division. Two ways are different if there is a position with a room divider installed in the first way but not in the second way.
Return the number of ways to divide the corridor. Since the answer may be very large, return it modulo 109 + 7. If there is no way, return 0.
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.length1 <= n <= 105corridor[i] is either 'S' or 'P'.This approach involves counting the total number of 'S' in the string. We need to ensure that the total number of seats is even, as every section must have exactly two seats. If the count of 'S' is not even or less than two, return 0 because no valid sections can be formed.
Initialize a counter for seats and iterate through the corridor. Each time you encounter the second seat in a pair, check how many plants were between this second seat and the previous pair of seats. The count of positions (plants) between these pairs is the spot where dividers can be placed. Compute the total number of ways combine these choices.
This implementation first counts all 'S' in the corridor to determine if it's possible to split the corridor. It then iterates through the string, calculates how many plants ('P') are between every two pairs of seats ('S'), and uses this count to determine the number of ways dividers can be placed.
JavaScript
Time Complexity: O(n), where n is the length of the corridor string because it processes the string in a single pass.
Space Complexity: O(1), since it uses a fixed amount of extra space for variables.
The two-pointer technique helps efficiently track the segments of the corridor. Position both pointers initially at the start of the string. Move the first pointer until two 'S' characters are found, marking the start of a section. Move the second pointer to continue finding the next two 'S' for another section.
Each transition between segments of groups discovered by the pointers represents a potential divider placement, and the number of these potential movements gives the number of possible configurations.
This program approaches the corridor string in one pass, using a variable to determine when to apply the mod operation for calculating potential dividers and calculates the result in a manner similar to counting plants between seat pairs.
Java
Time Complexity: O(n) due to the string being processed a single time.
Space Complexity: O(1), since only a fixed number of state-tracking variables are used.
| Approach | Complexity |
|---|---|
| Counting Seats and Calculating Divisions | Time Complexity: O(n), where n is the length of the corridor string because it processes the string in a single pass. Space Complexity: O(1), since it uses a fixed amount of extra space for variables. |
| Two-Pointer Technique | Time Complexity: O(n) due to the string being processed a single time. Space Complexity: O(1), since only a fixed number of state-tracking variables are used. |
Number of Ways to Divide a Long Corridor - Leetcode 2147 - Python • NeetCodeIO • 6,566 views views
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