<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 = {}
        n = len(word1)
        m = len(word2)

        def LCS(i,j):
            nonlocal n,m,word1,word2
            if i==n or j==m:
                return 0

            if (i,j) in cache:
                return cache[(i,j)]

            if word1[i]==word2[j]:
                cache[(i,j)] = 1 + LCS(i+1,j+1)
                return cache[(i,j)]

            cache[(i,j)] = max(LCS(i,j+1),LCS(i+1,j))

            return cache[(i,j)]

        lcs = LCS(0,0)

        return n+m - 2*lcs



        
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