<p>You are given a string <code>s</code> of length <code>n</code> where <code>s[i]</code> is either:</p>

<ul>
	<li><code>&#39;D&#39;</code> means decreasing, or</li>
	<li><code>&#39;I&#39;</code> means increasing.</li>
</ul>

<p>A permutation <code>perm</code> of <code>n + 1</code> integers of all the integers in the range <code>[0, n]</code> is called a <strong>valid permutation</strong> if for all valid <code>i</code>:</p>

<ul>
	<li>If <code>s[i] == &#39;D&#39;</code>, then <code>perm[i] &gt; perm[i + 1]</code>, and</li>
	<li>If <code>s[i] == &#39;I&#39;</code>, then <code>perm[i] &lt; perm[i + 1]</code>.</li>
</ul>

<p>Return <em>the number of <strong>valid permutations</strong> </em><code>perm</code>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;DID&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> The 5 valid permutations of (0, 1, 2, 3) are:
(1, 0, 3, 2)
(2, 0, 3, 1)
(2, 1, 3, 0)
(3, 0, 2, 1)
(3, 1, 2, 0)
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;D&quot;
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>n == s.length</code></li>
	<li><code>1 &lt;= n &lt;= 200</code></li>
	<li><code>s[i]</code> is either <code>&#39;I&#39;</code> or <code>&#39;D&#39;</code>.</li>
</ul>

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

    int numPermsDISequence(string s) {
        int n = s.size();
        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(n + 1, vector<int>(n + 1, 0)));

        // Base case: Only one permutation when there are no characters to process
        dp[0][0][0] = 1;

        for (int i = 1; i <= n; i++) {
            for (int greater = 0; greater <= i; greater++) {
                for (int smaller = 0; smaller <= i; smaller++) {
                    if (greater + smaller > i) continue;

                    int ans = 0;
                    if (s[i - 1] == 'D') {
                        // We need to place a smaller element than the last
                        for (int j = 0; j < smaller; j++) {
                            ans = (ans + dp[i - 1][greater + (smaller - j - 1)][j]) % mod;
                        }
                    } else {
                        // We need to place a greater element than the last
                        for (int j = 0; j < greater; j++) {
                            ans = (ans + dp[i - 1][j][smaller + (greater - j - 1)]) % mod;
                        }
                    }

                    dp[i][greater][smaller] = ans;
                }
            }
        }

        // Sum all valid permutations that can end with a valid configuration
        int result = 0;
        for (int greater = 0; greater <= n; greater++) {
            for (int smaller = 0; smaller <= n; smaller++) {
                result = (result + dp[n][greater][smaller]) % mod;
            }
        }

        return result;
    }
};

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