<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:

        m = len(word1)
        n = len(word2)

        dp = [ [ -1 for _ in range(n + 1) ] for _ in range(m + 1) ]

        def lcs(m, n):
            if m == 0 or n == 0:
                return ""

            if dp[m][n] != -1:
                return dp[m][n]

            if word1[m - 1] == word2[n - 1]:
                dp[m][n] = lcs(m - 1, n - 1) + word1[m - 1]
                return dp[m][n]
            else:
                choice1 = lcs(m, n - 1)
                choice2 = lcs(m - 1, n)
                dp[m][n] = choice1 if len(choice1) > len(choice2) else choice2
                return dp[m][n]

        res = lcs(m, n)
        print(res)
        return m + n - 2*len(res)

        
19698