# 2147. Number of Ways to Divide a Long Corridor

## Description

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

## Solutions

• class Solution {
private String s;
private int n;
private int[][] f;
private static final int MOD = (int) 1e9 + 7;

public int numberOfWays(String corridor) {
s = corridor;
n = s.length();
f = new int[n][3];
for (var e : f) {
Arrays.fill(e, -1);
}
return dfs(0, 0);
}

private int dfs(int i, int cnt) {
if (i == n) {
return cnt == 2 ? 1 : 0;
}
cnt += s.charAt(i) == 'S' ? 1 : 0;
if (cnt > 2) {
return 0;
}
if (f[i][cnt] != -1) {
return f[i][cnt];
}
int ans = dfs(i + 1, cnt);
if (cnt == 2) {
ans += dfs(i + 1, 0);
ans %= MOD;
}
f[i][cnt] = ans;
return ans;
}
}

• class Solution {
public:
const int mod = 1e9 + 7;

int numberOfWays(string corridor) {
int n = corridor.size();
vector<vector<int>> f(n, vector<int>(3, -1));
function<int(int, int)> dfs;
dfs = [&](int i, int cnt) {
if (i == n) return cnt == 2 ? 1 : 0;
cnt += corridor[i] == 'S';
if (cnt > 2) return 0;
if (f[i][cnt] != -1) return f[i][cnt];
int ans = dfs(i + 1, cnt);
if (cnt == 2) {
ans += dfs(i + 1, 0);
ans %= mod;
}
f[i][cnt] = ans;
return ans;
};
return dfs(0, 0);
}
};

• class Solution:
def numberOfWays(self, corridor: str) -> int:
@cache
def dfs(i, cnt):
if i == n:
return int(cnt == 2)
cnt += corridor[i] == 'S'
if cnt > 2:
return 0
ans = dfs(i + 1, cnt)
if cnt == 2:
ans += dfs(i + 1, 0)
ans %= mod
return ans

n = len(corridor)
mod = 10**9 + 7
ans = dfs(0, 0)
dfs.cache_clear()
return ans


• func numberOfWays(corridor string) int {
n := len(corridor)
var mod int = 1e9 + 7
f := make([][]int, n)
for i := range f {
f[i] = make([]int, 3)
for j := range f[i] {
f[i][j] = -1
}
}
var dfs func(i, cnt int) int
dfs = func(i, cnt int) int {
if i == n {
if cnt == 2 {
return 1
}
return 0
}
if corridor[i] == 'S' {
cnt++
}
if cnt > 2 {
return 0
}
if f[i][cnt] != -1 {
return f[i][cnt]
}
ans := dfs(i+1, cnt)
if cnt == 2 {
ans += dfs(i+1, 0)
ans %= mod
}
f[i][cnt] = ans
return ans
}
return dfs(0, 0)
}

• function numberOfWays(corridor: string): number {
const M: number = 1e9 + 7;
const seatNumbers: number[] = [];

for (let i = 0; i < corridor.length; i++) {
if (corridor.charAt(i) === 'S') {
seatNumbers.push(i);
}
}

if (seatNumbers.length % 2 !== 0 || seatNumbers.length === 0) {
return 0;
}

let result: number = 1;

for (let i = 2; i < seatNumbers.length; i += 2) {
result = (result * (seatNumbers[i] - seatNumbers[i - 1])) % M;
}

return result;
}