<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>

class Solution:
    def numberOfWays(self, corridor: str) -> int:
        # Store 1000000007 in a variable for convenience
        MOD = 1_000_000_007

        # Cache the result of each sub-problem
        cache = [[-1] * 3 for _ in range(len(corridor))]

        # Count the number of ways to divide from "index" to the last index
        # with "seats" number of "S" in the current section
        def count(index, seats):
            # If we have reached the end of the corridor, then
            # the current section is valid only if "seats" is 2
            if index == len(corridor):
                return 1 if seats == 2 else 0

            # If we have already computed the result of this sub-problem,
            # then return the cached result
            if cache[index][seats] != -1:
                return cache[index][seats]
            
            # If the current section has exactly 2 "S"
            if seats == 2:
                # If the current element is "S", then we have to close the
                # section and start a new section from this index. Next index
                # will have one "S" in the current section
                if corridor[index] == "S":
                    result = count(index + 1, 1)
                else:
                    # If the current element is "P", then we have two options
                    # 1. Close the section and start a new section from this index
                    # 2. Keep growing the section
                    result = (count(index + 1, 0) + count(index + 1, 2)) % MOD
            else:
                # Keep growing the section. Increment "seats" if present
                # element is "S"
                if corridor[index] == "S":
                    result = count(index + 1, seats + 1)
                else:
                    result = count(index + 1, seats)
            
            # Memoize the result, and return it
            cache[index][seats] = result
            return cache[index][seats]
        
        # Call the count function
        return count(0, 0)
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