<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 lcs(i, j, s1, s2):
            if i < 0 or j < 0:
                return 0

            if (i, j) in memo:
                return memo[(i, j)]
            if s1[i] == s2[j]:
                memo[(i, j)] = 1 + lcs(i - 1, j - 1,s1, s2)
            else:
                memo[(i, j)] = max(lcs(i - 1, j, s1, s2), lcs(i, j - 1, s1, s2))

            return memo[(i, j)]
        
        c = lcs(len(word1)-1, len(word2)-1, word1, word2)
        print(c)
        return len(word1) - c + len(word2) - c
        
29936