<p>Given two strings <code>word1</code> and <code>word2</code>, return <em>the minimum number of <strong>steps</strong> required to make</em> <code>word1</code> <em>and</em> <code>word2</code> <em>the same</em>.</p>

<p>In one <strong>step</strong>, you can delete exactly one character in either string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;sea&quot;, word2 = &quot;eat&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You need one step to make &quot;sea&quot; to &quot;ea&quot; and another step to make &quot;eat&quot; to &quot;ea&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;leetcode&quot;, word2 = &quot;etco&quot;
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= word1.length, word2.length &lt;= 500</code></li>
	<li><code>word1</code> and <code>word2</code> consist of only lowercase English letters.</li>
</ul>

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        memo = {}
        def dfs(pw1, pw2):
            if pw1 >= len(word1):
                return len(word2) - pw2
            
            if pw2 >= len(word2):
                return len(word1) - pw1

            if (pw1, pw2) in memo:
                return memo[(pw1, pw2)]

            if word1[pw1] == word2[pw2]:
                return dfs(pw1 + 1, pw2 + 1)
            
            elif word1[pw1] != word2[pw2]:
                path1 = 1 + dfs(pw1 + 1, pw2)
                path2 = 1 + dfs(pw1, pw2 + 1)
                memo[(pw1, pw2)] = min(path1, path2)
                return memo[(pw1, pw2)]

        return dfs(0,0)
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