<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:
        n = len(s)
        # for round s[:j], starting at j=0
        # dp[i] = # of valid permutations that ends with i
        # at round j s[:j], we only have dp[0...j] positions filled for valid permutations
        dp = [0 for i in range(n+1)]
        dp[0] = 1
        MOD = 10 ** 9 + 7

        for i in range(n):
            c = s[i]
            new_dp = [0 for i in range(n+1)]
            prefix_sum = 0
            if c == 'I':
                # _...k + j where 0 <= k <= j-1
                # new_dp[j] = dp[0] + dp[1] + ... + dp[j-1]
                for j in range(i+2):
                    new_dp[j] = prefix_sum
                    prefix_sum = (prefix_sum + dp[j]) % MOD
            elif c == 'D':
                # _...k + j where j <= k <= i
                # new_dp[j] = dp[j] + dp[j+1] + ... dp[n]
                for j in range(i+1, -1, -1):
                    prefix_sum = (prefix_sum + dp[j]) % MOD
                    new_dp[j] = prefix_sum
            dp = new_dp
        print(dp)
        return sum(dp) % MOD
16500