<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:
    def numPermsDISequence(self, s: str) -> int:
        """
        1) brute force: all possible perms and check if valid => n+1!*n (too big) 
        2) dynamic programming: since if we are at an idx i, the rest of the substring is getting repeated.
        state => (current idx of s, no. choices above prev, no. choices below prev)
        transition => case1: 'D' we can choose any element from above choices. 
                      case2: 'I' we can choose any element from below choices.
        """
        n = len(s)
        mod = 10**9+7
        ans = 0 
        @cache
        def rec(idx, a, b):
            # if it reaches end it means its a valid sequence (implicitly means a=0, b=0)
            if idx == n: return 1   
            # initialize valid sequences to 0 
            ret = 0 
            if s[idx] == 'D':
                # if current idx is D, then we choose any element from the 'b' elements below it.
                # correspondingly update a and b (elements above and below element we choose)
                for i in range(1,b+1):
                    ret+=rec(idx+1, a+i-1, b-i)
            elif s[idx] == 'I':
                for i in range(1,a+1):
                    ret+=rec(idx+1, a-i, b+i-1)
            return ret%mod
        # we take first element as any of the 0 to n. and find for all starting elements. 
        for i in range(n+1):
            ans = (ans + rec(0, n-i, i))%mod
        return ans
25500