<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 dp[505][505];
int solve(int i,int j,string word1, string word2){
    if(i==word1.size()) return word2.size()-j;
    if(j==word2.size()) return word1.size()-i;
    if(dp[i][j]!=-1) return dp[i][j];
    if(word1[i]==word2[j]) return dp[i][j]= solve(i+1,j+1,word1,word2);
    return dp[i][j]=min({1+solve(i+1,j,word1,word2),1+solve(i,j+1,word1,word2)});
}
    int minDistance(string word1, string word2) {
        memset(dp,-1,sizeof dp);
        return solve(0,0,word1,word2);
    }
};
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