<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>

typedef long long ll;
class Solution {
public:
    int e=1e9+7;
    vector<vector<ll>>dp;
    vector<int>vis;
    int solve(int ind,int prev,string &s)
    {
        int n=s.size();
        if(ind==n){
            return 1;
        }
        if(dp[prev][ind]!=-1) return dp[prev][ind];
        ll ans=0;
        for(int i=0;i<=n;i++)
        {
            if(vis[i]) continue;
            if(s[ind]=='D'&&i>prev) continue;
            if(s[ind]=='I'&&i<prev) continue;
            vis[i]=1;
            ans=(ans+solve(ind+1,i,s))%e;
            vis[i]=0;
        }
        return dp[prev][ind]=ans;
    }
    int numPermsDISequence(string s) {
        int n=s.size();
        dp.resize(n+1,vector<ll>(n+1,-1));
        vis.resize(n+1,0);
        int ans=0;
        for(int i=0;i<=n;i++)
        {
            vis[i]=1;
            ans=(ans+solve(0,i,s))%e;
            vis[i]=0;
        }
        return ans;
    }
};
10300