<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 {
private:
    int mod = 7 + 1e9;
    vector<vector<int>> dp;

    int valid_perm(int ind, string& s, int last, vector<bool>& vis) {
        if (ind == s.length() + 1) { 
            return 1;
        }

        if (dp[ind][last] != -1) {
            return dp[ind][last];
        }

        int cursum = 0;

        for (int i = 0; i <= s.length(); i++) {
            if (!vis[i]) {  
                if (s[ind - 1] == 'D') {
                    if (last > i) {
                        vis[i] = true;
                        cursum += valid_perm(ind + 1, s, i, vis); 
                        cursum %= mod;
                        vis[i] = false;
                    }
                } 
                else {
                    if (last < i) {
                        vis[i] = true;
                        cursum += valid_perm(ind + 1, s, i, vis); 
                        cursum %= mod;
                        vis[i] = false;
                    }
                }
            }
        }

        return dp[ind][last] = cursum % mod;
    }

public:
    int numPermsDISequence(string s) {
        int n = s.length();
        dp = vector<vector<int>> (n + 1, vector<int>(n + 1, -1));
        vector<bool> vis(n + 1, false);

        int sum = 0;
        for (int i = 0; i <= n; i++) {
            vis[i] = true;
            sum += valid_perm(1, s, i, vis);  
            sum %= mod;
            vis[i] = false;
        }
        return sum % mod;
    }
};
10400