<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:
    def numDistinct(self, s: str, t: str) -> int:

        n = len(s)
        m = len(t)

        dp = [[0]*(m + 1) for _ in range(n + 1)]

        for i in range(1, n + 1):
            for j in range(1, m + 1):
                if s[i - 1] == t[j - 1]:
                    dp[i][j] = dp[i-1][j -1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][ j - 1])

        if dp[n][m] != m:
            return 0

        seen = [[0]*(m + 1) for _ in range(n + 1)]
        
        def solve(i, j):

            if seen[i][j]:
                return seen[i][j]
            
            if i == 0 or j == 0:
                return 1
            
            left = dp[i][j -1] 
            top = dp[i - 1][j]
            diag = dp[i - 1][j - 1]

            if dp[i][j] == diag + 1 and s[i - 1] == t[j - 1] :
                seen[i][j] = solve(i - 1, j - 1)
            if left == dp[i][j]:
                seen[i][j] = seen[i][j] + solve(i, j - 1)
            if top == dp[i][j]:
                seen[i][j] = seen[i][j] + solve(i - 1, j)

            return seen[i][j]
        
        solve(n , m)
        
        return seen[n][m]
        
46605