<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:
    const int d=1e9+7;
    int numPermsDISequence(string s) {
        int n=s.size();
        int dp[205][205]={},pre[205][205]={};
        dp[0][0]=1;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n+1;j++){
                pre[i-1][j]=pre[i-1][j-1]+dp[i-1][j-1];
                pre[i-1][j]%=d;
            }
            for(int j=0;j<=i;j++){
                if(s[i-1]=='I'){
                    dp[i][j]+=pre[i-1][j];
                    dp[i][j]%=d;
                }
                else{
                    if(pre[i-1][i]<pre[i-1][j]) pre[i-1][i]+=d;
                    dp[i][j]+=pre[i-1][i]-pre[i-1][j];
                    dp[i][j]%=d;
                }
            }
        }
        int ans=0;
        for(int i=0;i<=n;i++){
            ans+=dp[n][i];
            ans%=d;
        }
        return ans;
    }
};
8100