<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:
        i = 0 # t string
        j = 0 #s string
        m = len(t)
        n = len(s)
        nextoccurance = defaultdict(list)
        for i,a in enumerate(s):
            nextoccurance[a].append(i)
        def bs(arr,target):
            f = 0
            l =len(arr)-1
            res = -1
            while f<=l:
                mid = f + (l-f)//2
                if arr[mid]>=target:
                    res = mid
                    l = mid -1
                else:
                    f = mid+1
            if res == len(arr)-1:
                return -1
            return arr[res+1]
        @cache
        def dp(i,j):
            #print(i,j)
            if i==m:
                return 1
            if j==n:
                return 0
            if s[j]==t[i]:
                nextocc =bs(nextoccurance[t[i]],j)
                if nextocc ==-1:
                    val = dp(i+1,j+1)
                else:
                    val = dp(i+1,j+1) + dp(i,nextocc)
            else:
                val = dp(i,j+1)
            #print(i,j,val)
            return val
        return dp(0,0)
            

        
124555