<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 solve(string text1, string text2,int i, int j,vector<vector<int>> &dp){
        if(i == text1.length()){
            return 0;
        }
        if(j == text2.length()){
            return 0;
        }
        if(dp[i][j] != -1){
            return dp[i][j];
        }
        int ans = 0;
        if(text1[i] == text2[j]){
            ans = 1 + solve(text1,text2,i+1,j+1,dp);
        }
        else{
             ans = max(solve(text1,text2,i+1,j,dp),solve(text1,text2,i,j+1,dp));
        }
        return dp[i][j] = ans; 
    }
    int minDistance(string text1, string text2) {
        vector<vector<int>> dp(text1.size(),vector<int>(text2.size(),-1));
       int k = solve(text1,text2,0,0,dp); 
       return (text1.size()-k)+(text2.size()-k);
    }
};
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