<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 {
    int func_dp(string &text1,string &text2,int index1,int index2,vector<vector<int>> &dp)
    {
        if(index1==text1.size() || index2==text2.size()) return 0;
        if(dp[index1][index2]!=-1) return dp[index1][index2];
        
        if(text1[index1] == text2[index2]) return dp[index1][index2] = 1 + func_dp(text1,text2,index1+1,index2+1,dp);

        return dp[index1][index2] = 0 + max(func_dp(text1,text2,index1,index2+1,dp) , func_dp(text1,text2,index1+1,index2,dp));
        
    }
public:
    int minDistance(string word1, string word2) {
        int n = word1.size();
        int m = word2.size();

        vector<vector<int>> dp(n+1,vector<int>(1000+1,-1));

        int x = func_dp(word1,word2,0,0,dp);
        return ((n-x) + (m-x));


    }
};
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