<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:

    # top down with memoization
    def minDistance_(self, word1: str, word2: str) -> int:
        
        cache = {}

        def dfs(i, j):
            # iterative case
            if (i, j) in cache:
                return cache[(i, j)]
            elif i < len(word1) and j < len(word2):
                # progress with cost of 0
                if word1[i] == word2[j]:
                    cache[(i, j)] = dfs(i+1, j+1)
                # 3 operations each with a cost of 1
                else:
                    cache[(i, j)] = 1 + min(dfs(i, j+1), dfs(i+1, j), dfs(i+1, j+1))
                return cache[(i, j)]
            # base case
            else:
                return max(len(word1)-i, len(word2)-j)
        
        return dfs(0, 0)
    

    def minDistance(self, word1: str, word2: str) -> int:
        row  = [i for i in reversed(range(len(word2)+1))]
        for i in reversed(range(len(word1))):
            newrow = [None for _ in range(len(word2)+1)]
            newrow[-1] = len(word1)-i
            for j in reversed(range(len(word2))):
                if word1[i] == word2[j]:
                    newrow[j] = row[j+1]
                else:
                    newrow[j] = 1 + min(newrow[j+1], row[j+1], row[j])
            row = newrow
        return row[0]
16762