<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, n = len(word1), len(word2)

        def lcs(s1, s2, m, n):
            if m == 0 or n == 0:
                return 0

            if s1[m - 1] == s2[n - 1]:
                return 1 + lcs(s1, s2, m - 1, n - 1)
            else:
                return max(lcs(s1, s2, m - 1, n), lcs(s1, s2, m, n - 1))

        def lcs_memo(s1, s2, m, n, memo=None):
            if memo is None:
                memo = {}

            if m == 0 or n == 0:
                return 0

            if (m, n) in memo:
                return memo[(m, n)]

            if s1[m - 1] == s2[n - 1]:
                memo[(m, n)] = 1 + lcs_memo(s1, s2, m - 1, n - 1, memo)
            else:
                memo[(m, n)] = max(lcs_memo(s1, s2, m - 1, n, memo), lcs_memo(s1, s2, m, n - 1, memo))

            return memo[(m, n)]

        return m + n - lcs_memo(word1, word2, m, n) * 2
17878