<p>Given two strings s and t, return <i>the number of distinct</i> <b><i>subsequences</i></b><i> of </i>s<i> which equals </i>t.</p>

<p>The test cases are generated so that the answer fits on a 32-bit signed integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;rabbbit&quot;, t = &quot;rabbit&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong>
As shown below, there are 3 ways you can generate &quot;rabbit&quot; from s.
<code><strong><u>rabb</u></strong>b<strong><u>it</u></strong></code>
<code><strong><u>ra</u></strong>b<strong><u>bbit</u></strong></code>
<code><strong><u>rab</u></strong>b<strong><u>bit</u></strong></code>
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;babgbag&quot;, t = &quot;bag&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong>
As shown below, there are 5 ways you can generate &quot;bag&quot; from s.
<code><strong><u>ba</u></strong>b<u><strong>g</strong></u>bag</code>
<code><strong><u>ba</u></strong>bgba<strong><u>g</u></strong></code>
<code><u><strong>b</strong></u>abgb<strong><u>ag</u></strong></code>
<code>ba<u><strong>b</strong></u>gb<u><strong>ag</strong></u></code>
<code>babg<strong><u>bag</u></strong></code></pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length, t.length &lt;= 1000</code></li>
	<li><code>s</code> and <code>t</code> consist of English letters.</li>
</ul>

class Solution {
public:
    int numDistinct(string s, string t) {
        int n = s.size();
        int m = t.size();
        if(n < m) return 0;
        vector<int> prev(m+1, 0);
        vector<long long int> prevn(m+1, 1);

        for(int i = 1; i <= n; i++){
            vector<int> cur(m+1, 0);
            vector<long long int> curn(m+1, 0);
            curn[0] = 1;
            for(int j = 1; j <= m; j++){
                int a = prev[j];
                int b =  cur[j-1];
                if( a == b){
                    curn[j] = (prevn[j] + curn[j-1]) % 100000000009;
                }else if( a > b){
                    curn[j] = prevn[j] % 100000000009;
                }else{
                    curn[j] = curn[j-1] % 100000000009;
                    a = b;
                }
                if(s[i-1] == t[j-1]){
                        b = 1 + prev[j-1]; 
                }
                
                if( a == b){
                    curn[j] += (prevn[j-1]) % 100000000009;
                    cur[j] = a;
                }else if( a < b){
                    curn[j] = prevn[j-1] % 1000000009;
                    cur[j] = b;
                }else{
                    cur[j] = a;
                }

            }
            prev = cur;
            prevn = curn;
        }
        
        if(prev[m] == 0 || prev[m] != m) return 0;
        return prevn[m];
    }
};
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