<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:
        #here we need to return the min no of deletions 
        m=len(word1)
        n=len(word2)
        dp=[[-1 for i in range(n+1)]for j in range(m+1)]
        a=m-self.helper(m-1,n-1,word1,word2,m,n,dp)
        b=n-self.helper(m-1,n-1,word1,word2,m,n,dp)
        return (a+b)

    def helper(self,ind1,ind2,s1,s2,m,n,dp):
        if(ind1<0 or ind2<0):
            return 0
        if(dp[ind1][ind2]!=-1):
            return dp[ind1][ind2]
        if(s1[ind1]==s2[ind2]):
            return 1+self.helper(ind1-1,ind2-1,s1,s2,m,n,dp)
        else:
            ele1=self.helper(ind1,ind2-1,s1,s2,m,n,dp)
            ele2=self.helper(ind1-1,ind2,s1,s2,m,n,dp)
            dp[ind1][ind2]=max(ele1,ele2)
            return dp[ind1][ind2]
       
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