<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 = dict()
        def dp(one, two, oneStr, twoStr): # min num deletion 
            if (one, two) in cache:
                return cache[(one, two)]
            if one == len(oneStr) and two == len(twoStr):
                return 0
            if one == len(oneStr):
                return len(twoStr) - two
            if two == len(twoStr):
                return len(oneStr) - one
            
            res = None
            if oneStr[one] == twoStr[two]:
                res = dp(one+1, two+1,oneStr,twoStr)
            else:
                res = min(min(dp(one+1, two,oneStr,twoStr), dp(one, two+1,oneStr,twoStr)) + 1, dp(one+1,two+1,oneStr,twoStr) + 2)
            cache[(one, two)] = res
            
            return res

        return dp(0, 0, word1, word2)



        
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