<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 pprint import pprint

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        memo = {}
        def recursion(word,index):
            if word in memo:
                return memo[word]

            while index < len(word2) and index < len(word) and word[index] == word2[index]:
                index += 1
            
            if index == len(word2): 
                memo[word] = len(word) - len(word2) 
                return memo[word]
            
            if index == len(word):
                memo[word] = len(word2) - len(word) 
                return memo[word]                

            #index < len(word2)
            # can either replace or remove element, you can't add, adding will lead to 
            # an infinity of solutions. no it doesn't, if you add the corresponding 
            # letter eventually you will end up at the target index.
            res1 = 1 + recursion(word[:index]+word[index + 1:],index)            
            res2 = 1 + recursion(word[:index]+word2[index]+word[index + 1:],index)            
            res3 = 1 + recursion(word[:index]+word2[index]+word[index:],index)            
            memo[word] = min(min(res1,res2) , res3)
            return memo[word]

        string = ""
        res = recursion(word1,0)
        return res






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