<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:
    vector<bool>vis;
    long long mod = 1e9+7;
    vector<vector<int>>dp;
    long long func(string s, int idx, int prev){
      if(idx>=s.size())return 1;
      if(dp[idx][prev+1]!=-1)return dp[idx][prev+1];
      int co = 0;
      for(int i = 0;i<=s.size();i++){
        if(!vis[i]){
          if(prev==-1){
            vis[i] = true;
            co  = (co + func(s,idx,i))%mod;
          }
          else if((s[idx]=='D'&&i<prev)||(s[idx]=='I'&&i>prev)){
            
                vis[i] = true;
                 co  = (co + func(s,idx+1,i))%mod;
            
          }
          vis[i] = false;
      }
       
    }
    return dp[idx][prev+1] = co;
    }
    int numPermsDISequence(string s) {
         vis.resize(s.size()+1);
         dp.resize(s.size(),vector<int>(202,-1));
         return func(s,0,-1);
    }
};
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