<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 {
public:
 int f(string s1,string s2,int n,int m,int i,int j,vector<vector<int>>&dp)
{ int len=0;
if(i<0||j<0) return len=0;
if(dp[i][j]!=-1) return dp[i][j];
if(s1[i]==s2[j]) len=1+f(s1,s2,n,m,i-1,j-1,dp);
else len=max(f(s1,s2,n,m,i,j-1,dp),f(s1,s2,n,m,i-1,j,dp));

return  dp[i][j]=len;
}

    int minDistance(string word1, string word2) {
    int n=word1.length();
    int m=word2.length();
    vector<vector<int>>dp(n+1,vector<int>(m+1,-1));
    return (n+m)-2*f(word1,word2,n,m,n-1,m-1,dp);   
    }
};
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