<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>

const int mod = 1e9+7;
class Solution {
    int n, vis[202], dp[202][202];
private:
    int f(int prev, int len, string& s){
        if(len == n+1) return 1;
        if(dp[prev+1][len] != -1) return dp[prev+1][len];
        int ans = 0;
        if(prev == -1){
            for(int i=0; i<=n; i++){
                vis[i] = 1;
                ans = (ans + f(i,len+1,s))%mod;
                vis[i] = 0;
            }
        }
        else if(prev != -1 && s[len-1] == 'D'){
            for(int i=0; i<prev; i++){
                if(!vis[i]){
                    vis[i] = 1;
                    ans = (ans + f(i,len+1,s))%mod;
                    vis[i] = 0;
                }
            }
        }
        else if(prev != -1 && s[len-1] == 'I'){
            for(int i=prev+1; i<=n; i++){
                if(!vis[i]){
                    vis[i] = 1;
                    ans = (ans + f(i,len+1,s))%mod;
                    vis[i] = 0;
                }
            }
        }

        return dp[prev+1][len] = ans;
    }
public:
    int numPermsDISequence(string s) {
        n = s.size();
        memset(vis, 0, sizeof(vis));
        memset(dp, -1, sizeof(dp));
        return f(-1,0,s);
    }
};
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