<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) {
        unordered_map<string, int> cache;
        return dfs(0, 0, s, t, cache);

    }

private: 
    int dfs(int i, int j, string& s, string& t,  unordered_map<string, int>& cache){
        if(j == t.length()){
            return 1; 
        }
        if(i == s.length()){
            return 0; 
        }

        string key = to_string(i) + '#' + to_string(j); 

        if(cache.count(key)){
            return cache[key]; 
        }
        int res = 0; 
        if(s[i] == t[j]){
            res = dfs(i +1, j+1, s, t, cache) + dfs(i+1, j, s, t, cache);
        } else{
            res = dfs(i+1, j, s, t, cache);
        }

        cache[key] = res; 

        return res;

    }
};
144803