<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>

//2 dp states i and j
//dp[i][j] represents the number of possibilities till index i with j+1 smallest number as the last bit
//Here j the smallest means j th smallest of the unused numbers 
#define ll long long
#define pb push_back
#define MOD 1000000007
class Solution {
public:
    ll f(ll less,ll more,vector<vector<ll>>& dp,ll n,string & s){
        ll index = n-(less+more);
        if(dp[less][more]!=-1){
            return dp[less][more];
        }
        // cout<<index<<endl;
        if(index==n){
            return dp[less][more]=1;
        }
        ll tot = 0;
        if(s[index]=='D'){
            for(ll i=0;i<less;i++){
                tot = (tot + f(i,(less+more)-1-i,dp,n,s))%MOD;
            }
        }
        else{
            for(ll i=0;i<more;i++){
                tot = (tot + f((less+more)-1-i,i,dp,n,s))%MOD;
            }
        }

        return dp[less][more]=tot;
    }
    int numPermsDISequence(string s) {

        ll n = s.size();
        vector<vector<ll>>dp(n+1,vector<ll>(n+1,-1));

        ll tot = 0;
        for(ll i=0;i<=n;i++){
            tot = (tot + f(i,n-i,dp,n,s))%MOD;
        }
        return tot;
    }
};
10000