<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:
        
        self.myDict = {}

        def recur(i, j):
            
            if (i, j) in self.myDict:
                return self.myDict[(i, j)]

            if i == len(word1) or j == len(word2):
                self.myDict[(i, j)] = 0
                return 0

            if word1[i] == word2[j]:
                a = recur(i + 1, j + 1) + 1
                self.myDict[(i, j)] = a 
                return a
            
            if word1[i] != word2[j]:
                
                b = max(recur(i + 1, j), recur(i, j + 1))
                self.myDict[(i, j)] = b
                return b

        commonWords = recur(0, 0)

        SizeOfRemovalFromWord1 = len(word1) - commonWords
        SizeOfAdditionToWord2 = len(word2) - commonWords

        return SizeOfRemovalFromWord1 + SizeOfAdditionToWord2

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