<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:
        
        # s= rabbbit
        #    i
        # t= rabbit
        #    j

        #    1
        # s= babgbag
        #    i
        # t= bag
        #     j
        # w=       1

        # at (i, j)

        #     4 if j == len(t), return 1
        #     3 if i == len(s), return 0
        #     1 if the chars are equal, check (i+1, j+1) and (i+1, j)
        #     2 if they are not equal, check (i+1, j)


        # time = space = O(len(s)*len(t))
        # @cache
        # def get_ways(i: int, j: int) -> int:
        #     if j < 0:
        #         return 1
        #     if i < 0:
        #         return 0
        
        #     ways = 0
        #     if s[i] == t[j]:
        #         ways += get_ways(i-1, j-1)
        #     ways += get_ways(i-1, j)
        #     return ways

        # return get_ways(len(s)-1, len(t)-1)

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

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

        for i in range(m+1):
            dp[i][0] = 1

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

        return dp[m][n]









        

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