<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:
        '''
            sea
            eat

            constraints: len(s1) != len(s2)

            longest common subsequence

            answer = max(len(s1), len(s2)) - len(longestSub)

            longest common subsequence

            s1 = sea
            s2 = eat
            s1[i] == s2[j]:
            return 1 + dfs(i+1, j+1)

            s1[i] != s2[j]
            return max(dfs(i, j+1), dfs(i+1, j))


        '''

        memo = {}

        def lcs(i, j):
            if (i, j) in memo:
                return memo[(i,j)]
            if i >= len(word1) or j >= len(word2):
                return 0
            mx = -1
            if word1[i] == word2[j]:
                mx = max(mx, 1 + lcs(i+1, j+1)) 

            else:
                mx = max(mx, lcs(i, j+1), lcs(i+1, j))

            memo[(i,j)] = mx

            return mx
        l = lcs(0, 0)
        return len(word1) - l + len(word2) - l
31756