<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 hlp(int i,int j, string t1,string t2,vector<vector<int>>& dp)
    {
        if(i==t1.size() || j==t2.size()) return 0;
        if(dp[i][j]!=-1) return dp[i][j];
        dp[i][j] = t1[i] == t2[j]? 1+hlp(i+1,j+1, t1,t2,dp):max(hlp(i,j+1,t1,t2,dp),hlp(i+1,j,t1,t2,dp));
        return dp[i][j];
    }
    int minDistance(string word1, string word2) {
        vector<vector<int>> dp(word1.size(),vector<int>(word2.size(),-1));
        int lcs = hlp(0,0,word1,word2,dp);

        return word1.size()+word2.size() -lcs-lcs;
    }
};
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