<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>

# horse
# ros

#idx1 = len(horse)
#idx2 = len(ros)

#idx1 = len(intention)
#idx2 = len(execution)

# if both chars are equal: just decrement idx1 and idx2
# if delete: decrement idx1 and noperations++
# if add: decrement idx2 and noperations++
# if replace: decrement idx1 and idx2 and noperations++

# word1 = o
# word2 = horse 

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

        @lru_cache 
        def operation(idx1: int, idx2: int) -> int:
            if not len(word1):
                return len(word2)
            elif not len(word2):
                return len(word1)


            if idx1 == len(word1) and idx2 == len(word2):
                return 0
            else:
                if idx1 == len(word1):
                    return (len(word2)-idx2)
                elif idx2 == len(word2):
                    return (len(word1)-idx1)
                elif word1[idx1] == word2[idx2]:
                    return operation(idx1+1, idx2+1)
                else:
                    delete = operation(idx1+1, idx2) + 1
                    add = operation(idx1, idx2+1) + 1
                    replace = operation(idx1+1, idx2+1) + 1

                    return min(delete, add, replace)
        
        return operation(0, 0)
                
                
                    


17637