<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;
    long long dp[202][202];
    long long perms(string &s, int n, int pos, int prev, vector<bool> &visited) {
        if(pos >= n) return 1; 
        if(dp[pos][prev] != -1) return dp[pos][prev];
        long long ans = 0;
        for(int i=0;i<=n;i++) {
            if(!visited[i]) {
                if(s[pos] == 'D' && prev > i) {
                    visited[i] = true;
                    ans = (ans + perms(s,n,pos+1,i,visited) % MOD) % MOD;
                    visited[i] = false;
                }
                if(s[pos] == 'I' && prev < i) {
                    visited[i] = true;
                    ans = (ans + perms(s,n,pos+1,i,visited) % MOD) % MOD;
                    visited[i] = false;
                }
            }
        }
        return dp[pos][prev] = ans;
    }
    int numPermsDISequence(string s) {
        int n = s.length();
        long long ans = 0;
        memset(dp,-1,sizeof(dp));
        for(int i=0;i<=n;i++) {
            vector<bool> visited(n+1,false);
            visited[i] = true;
            ans = (ans + perms(s,n,0,i,visited) % MOD) % MOD;
        }
        return ans;
    }
};
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