<p>Given two strings <code>word1</code> and <code>word2</code>, return <em>the minimum number of operations required to convert <code>word1</code> to <code>word2</code></em>.</p>

<p>You have the following three operations permitted on a word:</p>

<ul>
	<li>Insert a character</li>
	<li>Delete a character</li>
	<li>Replace a character</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;horse&quot;, word2 = &quot;ros&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
horse -&gt; rorse (replace &#39;h&#39; with &#39;r&#39;)
rorse -&gt; rose (remove &#39;r&#39;)
rose -&gt; ros (remove &#39;e&#39;)
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;intention&quot;, word2 = &quot;execution&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> 
intention -&gt; inention (remove &#39;t&#39;)
inention -&gt; enention (replace &#39;i&#39; with &#39;e&#39;)
enention -&gt; exention (replace &#39;n&#39; with &#39;x&#39;)
exention -&gt; exection (replace &#39;n&#39; with &#39;c&#39;)
exection -&gt; execution (insert &#39;u&#39;)
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= word1.length, word2.length &lt;= 500</code></li>
	<li><code>word1</code> and <code>word2</code> consist of lowercase English letters.</li>
</ul>

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        cache = {}
        def dfs(s1: str, i: int, j: int) -> int:
            if s1 == word2:
                return 0
            if j == len(word2):
                return len(s1) - len(word2)
            if i == len(s1):
                return max(len(word2) - len(s1), 0)
            if (s1, i, j) in cache:
                return cache[(s1, i, j)]

            if s1[i] == word2[j]:
                cache[(s1, i, j)] = dfs(s1, i + 1, j + 1)
            else:
                s1_del = s1[:i] + s1[i + 1:]
                s1_ins = s1[:i] + word2[j] + s1[i:]
                s1_rep = s1[:i] + word2[j] + s1[i + 1:]
                cache[(s1, i, j)] = min(
                    1 + dfs(s1_del, i, j), 
                    1 + dfs(s1_ins, i + 1, j + 1),
                    1 + dfs(s1_rep, i + 1, j + 1)
                )
            return cache[(s1, i, j)]
        return dfs(word1, 0, 0)

25762