<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 solve(string word1,string word2,int i,int j,vector<vector<int>> &dp)
      {   
         if(i>=word1.size() || j>=word2.size())
         {
            return 0;
         } 
         if(dp[i][j]!=-1)
         {
            return dp[i][j];
         }


          if(word1[i]==word2[j])
          {
            return dp[i][j] = 1 +  solve(word1,word2,i+1,j+1,dp);
          }
          
          else
          {
            return dp[i][j] = max(solve(word1,word2,i,j+1,dp),solve(word1,word2,i+1,j,dp));
          }


      }
    
     int minDistance(string word1, string word2) {
      
      vector<vector<int>> dp(word1.size()+1,vector<int>(word2.size()+1,-1));
        return (word1.size() + word2.size() - (2*solve(word1,word2,0,0,dp))); 
    }
};
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