<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:
        cache = {}

        def e(i1, i2):
            if i1 == len(word1) and i2 == len(word2):
                return 0
            if (i1, i2) in cache:
                return cache[(i1, i2)]
            
            if i1 < len(word1) and i2 < len(word2) and word1[i1] == word2[i2]:
                return e(i1 + 1, i2 + 1)

            res = float('inf')
            if i1 < len(word1):
                res = min(res, 1 + e(i1 + 1, i2))
            if i2 < len(word2):
                res = min(res, 1 + e(i1, i2 + 1))
            
            cache[(i1, i2)] = res
            return res

        return e(0, 0)
27661