<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 M = 1e9+7;
    int dp[205][205];

    int solve(int prev, int i, string& s, vector<int>& vis) {
        if (prev < 0 || prev > s.length()) return 0;
        if (i == s.length()) return 1;
        if (dp[i][prev]!=-1) return dp[i][prev];

        int cnt = 0;

        if (s[i] == 'D') {
            for (int j = 0; j < prev; j++) {
                if (vis[j] == 0) {
                    vis[j] = 1;
                    cnt = (cnt + solve(j, i + 1, s, vis)) % M;
                    vis[j] = 0;
                }
            }
        } else {
            for (int j = prev + 1; j <= s.length(); j++) {
                if (vis[j] == 0) {
                    vis[j] = 1;
                    cnt = (cnt + solve(j, i + 1, s, vis)) % M;
                    vis[j] = 0;
                }
            }
        }

        return dp[i][prev] = cnt;
    }

    int numPermsDISequence(string s) {
        int n = s.length();
        int cnt = 0;
        memset(dp, -1, sizeof(dp));
        for (int i = 0; i <= n; i++) {
            vector<int> vis(202);
            vis[i] = 1;
            cnt = (cnt + solve(i, 0, s, vis)) % M;
            vis[i] = 0;
        }

        return cnt;
    }
};

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