<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:
        n = len(corridor)
        arr = []
        sofa = 0
        cnt = 0
        for i in range(n):
            if corridor[i]=='S':
                cnt+=1
                sofa+=1
                if sofa==1 and len(arr)!=0:
                    arr[-1][1] = i
            if sofa==2:
                arr.append([i, -1])
                sofa = 0

        if cnt%2==1 or cnt==0:
            return 0
        ans = 1
        print(arr)
        for i, j in arr:
            if j!=-1:
                ans*=(j-i) % 1_000_000_007


        return ans% 1_000_000_007
722