<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:
vector<vector<int>>dp;
int f(string word1,string word2,int i,int j,vector<vector<int>>&dp)
{
if(i==word1.size())return word2.size()-j;
if(j==word2.size())return word1.size()-i;
if(dp[i][j]!=-1)return dp[i][j];
if(word1[i]==word2[j])return dp[i][j]=f(word1,word2,i+1,j+1,dp);
return dp[i][j]=min(1+f(word1,word2,i+1,j,dp),1+f(word1,word2,i,j+1,dp));

}
    int minDistance(string word1, string word2) {
        int m=word1.size();
        int n=word2.size();
        dp.clear();
        dp.resize(m,vector<int>(n,-1));


        return f(word1,word2,0,0,dp);
        
    }
};
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