<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 dp[1001][1001];
    int solve(int i, int j, string &s, string &t)
    {
        if(j==0)
            return 1;
        if(i==0)
            return 0;
        if(dp[i][j]!=-1)
            return dp[i][j];
        if(s[i-1]==t[j-1])
            return dp[i][j]= solve(i-1, j-1, s, t)+solve(i-1, j, s, t);
        else
            return dp[i][j]= solve(i-1, j, s, t);

    }
    int numDistinct(string s, string t) {
        // memset(dp, -1, sizeof(dp));
        // return solve(s.size(), t.size(), s, t);

        int n=s.size(), m=t.size();
        vector<vector<double>> dp(n+1, vector<double>(m+1, 0));
        for(int i=0; i<=n; i++)
            dp[i][0]=1;
        for(int i=1; i<=m; i++)
            dp[0][i]=0;
        for(int i =1;i<=n;i++)
        {
            for(int j =1;j<=m;j++)
            {
                if(s[i-1]==t[j-1])
                    dp[i][j]=dp[i-1][j-1]+dp[i-1][j];
                else
                    dp[i][j]=dp[i-1][j];
            }
        } 
        return (int)dp[n][m];
    }
};
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