Number of Ways to Divide a Long Corridor
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2100-2199/2147.Number%20of%20Ways%20to%20Divide%20a%20Long%20Corridor/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2100-2199/2147.Number%20of%20Ways%20to%20Divide%20a%20Long%20Corridor/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2100-2199/2147.Number%20of%20Ways%20to%20Divide%20a%20Long%20Corridor/images/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>
Math; String; Dynamic Programming
function numberOfWays(corridor: string): number {
    const n = corridor.length;
    const mod = 10 ** 9 + 7;
    const f: number[][] = Array.from({ length: n }, () => Array(3).fill(-1));
    const dfs = (i: number, k: number): number => {
        if (i >= n) {
            return k === 2 ? 1 : 0;
        }
        if (f[i][k] !== -1) {
            return f[i][k];
        }
        if (corridor[i] === 'S') {
            ++k;
        }
        if (k > 2) {
            return (f[i][k] = 0);
        }
        f[i][k] = dfs(i + 1, k);
        if (k === 2) {
            f[i][k] = (f[i][k] + dfs(i + 1, 0)) % mod;
        }
        return f[i][k];
    };
    return dfs(0, 0);
}
