<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:
        # l1,l2=len(word1),len(word2)
        # dp=[[0]*(l2+1) for _ in range(l1+1)]
        # maxi=0
        # for i in range(l1):
        #     for j in range(l2):
        #         if word1[i]==word2[j]:
        #             dp[i][j]=1+dp[i-1][j-1]
        #         else:
        #             dp[i][j]=max(dp[i-1][j],dp[i][j-1])
        #         maxi=max(maxi,dp[i][j])
        # print(maxi)
        # return (l1-maxi)+(l2-maxi)
        
    
        has={}
        
        def lcs(m,n):
            
            if (m,n) in has:
                return has[(m,n)]
            if m==-1 or n==-1:
                return 0
            maxlen=0
            if word1[m]==word2[n]:
                return 1+lcs(m-1,n-1)
            has[(m,n)]=max(maxlen,lcs(m-1,n),lcs(m,n-1))   
            return has[(m,n)]

        l1,l2=len(word1),len(word2)
        maxi=lcs(l1-1,l2-1)
        return (l1-maxi) +(l2-maxi)
        
    
        
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