<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>

from functools import lru_cache

@lru_cache(None)  # Cache all results with no limit on cache size
def levenshtein_distance_recursive(s1, s2):
    # Base cases
    if len(s1) == 0:
        return len(s2)
    if len(s2) == 0:
        return len(s1)

    # If the last characters are the same, no operation is needed
    if s1[-1] == s2[-1]:
        return levenshtein_distance_recursive(s1[:-1], s2[:-1])

    # If the last characters are different, consider all possibilities
    # and take the minimum:
    return min(
        levenshtein_distance_recursive(s1, s2[:-1]),    # Insertion
        levenshtein_distance_recursive(s1[:-1], s2),    # Deletion
        levenshtein_distance_recursive(s1[:-1], s2[:-1]) # Substitution
    ) + 1

class Solution:
    distances = {}

    def lev(self, word1: str, word2: str, start1: int, start2: int) -> int:
        """Computes the Levenshtein distance between two strings, each starting at the respective index given."""
        cached_result = self.distances.get((start1, start2), None)

        if cached_result is not None:
            return cached_result

        if start2 == len(word2):
            distance = len(word1) - start1
        elif start1 == len(word1):
            distance = len(word2) - start2
        elif word1[start1] == word2[start2]:
            distance = self.lev(word1, word2, start1 + 1, start2 + 1)
        else:
            distance = 1 + min(
                self.lev(word1, word2, start1 + 1, start2),
                self.lev(word1, word2, start1, start2 + 1),
                self.lev(word1, word2, start1 + 1, start2 + 1),
            )

        self.distances[(start1, start2)] = distance

        return distance

    def minDistance(self, word1: str, word2: str) -> int:
        return levenshtein_distance_recursive(word1, word2)#self.lev(word1, word2, 0, 0)
25637