<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:
        dp = {}

        def LCS(n1, n2):
            if n1 == 0 or n2 == 0:
                return 0
            if (n1,n2) in dp:
                return dp[(n1,n2)]

            if word1[n1-1] == word2[n2-1]:
                dp[(n1,n2)] = LCS(n1-1, n2-1) + 1
                return dp[(n1,n2)]
            
            dp[(n1,n2)] = max(
                LCS(n1-1, n2),
                LCS(n1, n2-1),
            )
            return dp[(n1,n2)]

        
        
        lcs = LCS(len(word1), len(word2))
        return len(word1) + len(word2) - 2 * lcs


        
28798