<p>Along a long library corridor, there is a line of seats and decorative plants. You are given a <strong>0-indexed</strong> string <code>corridor</code> of length <code>n</code> consisting of letters <code>&#39;S&#39;</code> and <code>&#39;P&#39;</code> where each <code>&#39;S&#39;</code> represents a seat and each <code>&#39;P&#39;</code> represents a plant.</p>

<p>One room divider has <strong>already</strong> been installed to the left of index <code>0</code>, and <strong>another</strong> to the right of index <code>n - 1</code>. Additional room dividers can be installed. For each position between indices <code>i - 1</code> and <code>i</code> (<code>1 &lt;= i &lt;= n - 1</code>), at most one divider can be installed.</p>

<p>Divide the corridor into non-overlapping sections, where each section has <strong>exactly two seats</strong> with any number of plants. There may be multiple ways to perform the division. Two ways are <strong>different</strong> if there is a position with a room divider installed in the first way but not in the second way.</p>

<p>Return <em>the number of ways to divide the corridor</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>. If there is no way, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/04/1.png" style="width: 410px; height: 199px;" />
<pre>
<strong>Input:</strong> corridor = &quot;SSPPSPS&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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, <strong>each</strong> section has exactly <strong>two</strong> seats.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/04/2.png" style="width: 357px; height: 68px;" />
<pre>
<strong>Input:</strong> corridor = &quot;PPSPSP&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/12/3.png" style="width: 115px; height: 68px;" />
<pre>
<strong>Input:</strong> corridor = &quot;S&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no way to divide the corridor because there will always be a section that does not have exactly two seats.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == corridor.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>corridor[i]</code> is either <code>&#39;S&#39;</code> or <code>&#39;P&#39;</code>.</li>
</ul>

MOD = 10 ** 9 + 7

class Solution:
    def numberOfWays(self, corridor: str) -> int:
        n = len(corridor)
        cache = [[0] * 3 for _ in range(n + 1)]

        # def dfs(i, seats):
        #     if i > n - 1:
        #         return seats == 2

        #     if cache[i][seats] != -1:
        #         return cache[i][seats]

        #     if corridor[i] == 'S':
        #         res = dfs(i + 1, seats % 2 + 1) % MOD

        #     else:
        #         if seats < 2:
        #             res = dfs(i + 1, seats) % MOD
        #         else:
        #             res = dfs(i + 1, 0) + dfs(i + 1, 2) % MOD

        #     cache[i][seats] = res
        #     return res

        cache[-1][0], cache[-1][1], cache[-1][2] = 0, 0, 1

        for i in range(n - 1, -1, -1):
            for j in range(3):     
                if corridor[i] == 'S':  
                    cache[i][j] = cache[i + 1][j % 2 + 1] % MOD
                else:
                    if j < 2:
                        cache[i][j] = cache[i + 1][j] % MOD
                    else:
                        cache[i][j] = (cache[i + 1][0] + cache[i + 1][2]) % MOD


            
        return cache[0][0] % MOD

        # Time O(N) Space O(N)
        
32600