<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>

#define ll long long
#include <iostream>
class Solution {
public:
    ll dp[205][205] = {0};
    ll rangeSum[205][205] = {0};
    int numPermsDISequence(string s) {
        int n = s.size()+1;
        ll mod = 1000000007;
        for(int i=1;i<=n;i++){
            rangeSum[1][i] = 1;
        }
        dp[1][1] = 1;
        for(int i=2;i<=n;i++){
            for(int j=1;j<=i;j++){
                if(s[i-2] == 'I'){
                    // ll ans = 0;
                    // for(int x =1;x<j;x++){
                    //     ans += dp[i-1][x];
                    // }
                    ll ans = rangeSum[i-1][j-1];
                    dp[i][j] = ans%mod;
                } else {
                    // ll ans = 0;
                    // for(int x=j;x<=i;x++){
                    //     ans += dp[i-1][x];
                    // }
                    ll ans = rangeSum[i-1][i-1]-rangeSum[i-1][j-1];
                    dp[i][j] = ans%mod;
                }


                rangeSum[i][j] = rangeSum[i][j-1] + dp[i][j];
                rangeSum[i][j] %= mod;
            }
        }

        ll ans=0;
        for(int i=1;i<=n;i++){
            ans += dp[n][i];
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=i;j++){
                cout << dp[i][j] << " ";
            }
            cout << "\n";
        }

        for(int i=1;i<=n;i++){
            for(int j=1;j<=i;j++){
                cout << rangeSum[i][j] << " ";
            }
            cout << "\n";
        }

        ans%=mod;
        if(ans < 0) ans+=mod;
        return (ans);
    }


};
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