<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

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:

        @lru_cache(maxsize=(len(word1)+1) * (len(word2)+1))
        def solve(ptr1: int, ptr2: int) -> int:
            word1_ = word1[ptr1:]
            word2_ = word2[ptr2:]

            if not word1_ or not word2_:
                return len(word1_) + len(word2_)

            if word1_[0] == word2_[0]:
                return solve(ptr1+1, ptr2+1)
            else:
                r_candidate = 1 + solve(ptr1+1, ptr2+1)
                d_candidate = 1 + solve(ptr1, ptr2+1)
                i_candidate = 1 + solve(ptr1+1, ptr2)
                return min(r_candidate, d_candidate, i_candidate)

        return solve(0, 0)

        # # init the dp table
        # dp = [[None] * (len(word1) + 1) for _ in range(len(word2) + 1)]
        # for r in range(len(word2) + 1):
        #     dp[r][-1] = len(word2) - r

        # for c in range(len(word1) + 1):
        #     dp[-1][c] = len(word1) - c

        # for r in range(len(word2) - 1, -1, -1):
        #     for c in range(len(word1) - 1, -1, -1):
        #         if word1[c] == word2[r]:  # char match, do nothing
        #             dp[r][c] = dp[r+1][c+1]
        #         else:
        #             dp[r][c] = min(1 + dp[r+1][c+1], 1 + dp[r+1][c], 1 + dp[r][c+1])  # replace, insert, delete

        # # print(dp)
        # return dp[0][0]

19387