<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 common(self, w1, w2):
        if not w1 or not w2: return 0

        if (w1, w2) not in self.mem:
            if w1[0]==w2[0]:
                self.mem[(w1, w2)] = 1+self.common(w1[1:], w2[1:])
            else:
                self.mem[(w1, w2)] = max(self.common(w1[1:], w2), self.common(w1, w2[1:]))
        return self.mem[(w1, w2)]

    def minDistance(self, word1: str, word2: str) -> int:
        self.mem = {}
        com = self.common(word1, word2)
        ans = len(word1)+len(word2)-com*2
        return ans
21518