<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:
long long get_longestCommon_subseq(long long ind1,long long  ind2, vector<vector<long long>>&dp,string text1,string text2){
    if(ind1>=text1.size() || ind2>=text2.size()){
        return 0;
    }
    int result;
    if(dp[ind1][ind2]!=-1) return dp[ind1][ind2];
    if(text1[ind1]==text2[ind2]){
      result= 1 + get_longestCommon_subseq(ind1+1,ind2+1,dp,text1,text2);
    }
    else 
    result=max( get_longestCommon_subseq(ind1,ind2+1,dp,text1,text2), get_longestCommon_subseq(ind1+1,ind2,dp,text1,text2));
    dp[ind1][ind2]=result;
    return dp[ind1][ind2];
}
    int minDistance(string text1, string text2) {
         long long n=text1.size();
       long long m=text2.size();
        vector<vector<long long>>dp(n,vector<long long>(m+1,-1));
       return (n+m - 2*get_longestCommon_subseq(0,0,dp,text1,text2));
    }
};
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