<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 longestCommonSubsequence(string text1, string text2) {
        int a=text1.size(), b=text2.size();

        vector<vector<int>>dp(a+1, vector<int>(b+1, 0));
        vector<vector<char>>dir(a+1, vector<char>(b+1, 'u'));

        for(int i=0; i<=a; i++){
            dp[i][0]=0;
            dir[i][0]='u';
        }
        for(int i=0; i<=b; i++){
            dp[0][i]=0;
            dir[0][i]='l';
        }

        for(int i=1; i<=a; i++){
            for(int j=1; j<=b; j++){
                if(text1[i-1]==text2[j-1]){
                    dp[i][j]= 1+dp[i-1][j-1];
                    dir[i][j]='c';
                }
                else{
                    if(dp[i-1][j]< dp[i][j-1]){
                        dp[i][j]=dp[i][j-1];
                        dir[i][j]='l';
                    }else{
                        dp[i][j]=dp[i-1][j];
                        dir[i][j]='u';
                    }
                }
            }
        }
        string cmn="";
        int i=a, j=b;
        while(i>0 && j>0){
            if(dir[i][j]=='c'){
                cmn=text1[i-1]+cmn;
                i--;
                j--;
            }
            if(dir[i][j]=='u') i--;
            if(dir[i][j]=='l') j--;
        }
        return cmn.size();
    }

    int minDistance(string word1, string word2) {
        int i= longestCommonSubsequence(word1, word2);
        return word1.size()+word2.size()- 2*i;
    }
};
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