<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 func(string s, string cur,string t, int i,int j){
        if (i>=s.size()){
            if (cur==t) return dp[i][j]=1;
            return dp[i][j]=0;
        }
        if (dp[i][j]!=-1) return dp[i][j];
        int l1=0;
        if (s[i]==t[j])
        l1=func(s,cur+s[i],t,i+1,j+1);
        int l2=func(s,cur,t,i+1,j);
        return dp[i][j] = l1+l2;

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