<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 longestCommonSubsequence(self, text1: str, text2: str) -> int:
        memo = {}
        def lcs(i, j):
            if i == len(text1) or j == len(text2):
                return 0
            if (i, j) in memo:
                return memo[(i, j)]
            else:
                k = memo[(i, j)] = 1 + lcs(i + 1, j + 1) if text1[i] == text2[j] else max(lcs(i, j + 1), lcs(i + 1, j))
                return k
        return lcs(0, 0)

    def minDistance(self, word1: str, word2: str) -> int:
        return len(word1) + len(word2) - 2 * self.longestCommonSubsequence(word1, word2) 
28116