<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 dp[502][502];
    int lcs(int ind1,int ind2,string word1,string word2){
           if(ind1<0 ||ind2<0 ) return 0;
           if(dp[ind1][ind2]!=-1) return dp[ind1][ind2];
         if(word1[ind1]==word2[ind2]) return dp[ind1][ind2]=lcs(ind1-1,ind2-1,word1,word2)+1;

         return dp[ind1][ind2]=max(lcs(ind1-1,ind2,word1,word2),lcs(ind1,ind2-1,word1,word2));
    }
    int minDistance(string word1, string word2) {
        

        int n=word1.size();
        int m=word2.size();
         memset(dp,-1,sizeof(dp));
        return n+m-2*(lcs(n-1,m-1,word1,word2));
    }
};
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