<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 f(int ind,string &s,int last,set<int> &st,  vector<vector<int>> &dp){
    // base case
    if(ind>=s.length()){
        return 1;
    }
    if (dp[ind][last]!=-1){
        return dp[ind][last];
    }

int way=0;
    if (s[ind]=='D'){
        for (int i=0;i<last;i+=1){
            if (st.find(i)==st.end()){
                st.insert(i);
                way+=(f(ind+1,s,i,st,dp)%mod);
                way%=mod;
                st.erase(i);

            }
        }
    }
    else {
        for (int i=last+1;i<=s.length();i+=1){
            if (st.find(i)==st.end()){
            st.insert(i);
            way+=(f(ind+1,s,i,st,dp)%mod);
            way%=mod;
            st.erase(i);

            }

        }
     
    }
    return dp[ind][last]=way%mod;
}
    int numPermsDISequence(string s) {
        int n=s.length();
        if(n==1){
            return 1;
        }
        vector<vector<int>> dp(n+1,vector<int> (n+1,-1));
        int ans=0;
        set<int> st;
        for (int i=0;i<=n;i+=1){
            st.insert(i);
            ans=(ans%mod+f(0,s,i,st,dp)%mod)%mod;
            st.erase(i);
        }
        return ans%mod;
        
    }
};
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