<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:
        # Invoke the power of the void
        if not t:
            return 1
        if not s:
            return 0

        # Summon the cosmic alphabet
        ALPHABET_SIZE = 256
        
        # Forge the quantum lookup array
        next_occurrence = [[-1] * ALPHABET_SIZE for _ in range(len(s) + 1)]
        
        # Weave the temporal tapestry
        for i in range(len(s) - 1, -1, -1):
            next_occurrence[i] = next_occurrence[i + 1][:]
            next_occurrence[i][ord(s[i])] = i
        
        # Initialize the probability wave
        dp = [0] * (len(t) + 1)
        dp[0] = 1
        
        # Traverse the multiverse of subsequences
        for i in range(len(s)):
            for j in range(len(t) - 1, -1, -1):
                if s[i] == t[j]:
                    if j == len(t) - 1:
                        dp[j + 1] += dp[j]
                    else:
                        next_pos = next_occurrence[i + 1][ord(t[j + 1])]
                        if next_pos != -1:
                            dp[j + 1] += dp[j]
        
        # Return the cosmic count, contained within mortal bounds
        return dp[-1] % (10**9 + 7)
19323