<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 {
    int mod = 1e9+7;
public:
    int numPermsDISequence(string s) {
        /*
            dp[i][num] = number of DI sequence and the end = num

            1 2 4 3 <-3 => 12543
            1 2 4 3 <-2 => 12432 
            if(s[i]=='D')
                dp[i][num] = dp[i-1][num]
                           + dp[i-1][num+1]
                           + dp[i-1][num+2]
                           ...
                           + dp[i-1][i-1]

                dp[i][num] = dp[i-1][num] +dp[i][num+1]
            if(s[i]=='I')
                dp[i][num] = dp[i-1][num-1]
                           + dp[i-1][num-2]
                           ...
                           + dp[i-1][0]
                dp[i][num] = dp[i-1][num-1] + dp[i][num-1]
        */
        int n = s.size();
        int dp[n+1][n+1];
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for(int i=1; i<=n; i++){
            if(s[i-1]=='I'){
                for(int num=0; num<=i; num++){
                    dp[i][num] = num>0 ?(dp[i-1][num-1] + dp[i][num-1])%mod :0;
                }
            }
            else if(s[i-1]=='D'){
                for(int num=i; num>=0; num--){
                    dp[i][num] = (dp[i-1][num] + (num+1<=n ?dp[i][num+1] :0))%mod;
                }
            }
        }

        int cnt = 0;
        for(int i=0; i<=n; i++) cnt = (cnt + dp[n][i])%mod;
        return cnt;
    }
};
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