<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:
    #     m, n = len(s), 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 - 1][j]
    #             else:
    #                 dp[i][j] = dp[i - 1][j]

    #     return dp[m][n]
    def numDistinct(self, s: str, t: str):
        dp = {}

        def helper(i1, i2):
            if i2 < 0:
                return 1
            if i1 < 0:
                return 0
            if str(i1) + ":" + str(i2) in dp:
                return dp[str(i1) + ":" + str(i2)]
            if s[i1] == t[i2]:
                dp[str(i1) + ":" + str(i2)] = helper(i1 - 1, i2 - 1) + helper(i1 - 1, i2)
            else:
                dp[str(i1) + ":" + str(i2)] = helper(i1 - 1, i2)
            return dp[str(i1) + ":" + str(i2)]

        return helper(len(s) - 1, len(t) - 1)
114811