<p>Given two strings <code>word1</code> and <code>word2</code>, return <em>the minimum number of operations required to convert <code>word1</code> to <code>word2</code></em>.</p>

<p>You have the following three operations permitted on a word:</p>

<ul>
	<li>Insert a character</li>
	<li>Delete a character</li>
	<li>Replace a character</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;horse&quot;, word2 = &quot;ros&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
horse -&gt; rorse (replace &#39;h&#39; with &#39;r&#39;)
rorse -&gt; rose (remove &#39;r&#39;)
rose -&gt; ros (remove &#39;e&#39;)
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;intention&quot;, word2 = &quot;execution&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> 
intention -&gt; inention (remove &#39;t&#39;)
inention -&gt; enention (replace &#39;i&#39; with &#39;e&#39;)
enention -&gt; exention (replace &#39;n&#39; with &#39;x&#39;)
exention -&gt; exection (replace &#39;n&#39; with &#39;c&#39;)
exection -&gt; execution (insert &#39;u&#39;)
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= word1.length, word2.length &lt;= 500</code></li>
	<li><code>word1</code> and <code>word2</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    int minDistance(string word1, string word2) {
        memo.clear();
        return dfs(word1, word2);
    }
private:
    unordered_map<string, unordered_map<string, int>> memo;

    int dfs(const string& word1, const string& word2) {
        // Base Case: when one of the strings is empty
        if (word1.empty()) return word2.size();
        if (word2.empty()) return word1.size();

        // Check if the result is already in the memo
        if (memo[word1].count(word2)) return memo[word1][word2];

        // If the first characters match, skip them
        if (word1[0] == word2[0]) {
            return memo[word1][word2] = dfs(word1.substr(1), word2.substr(1));
        }

        // Calculate the minimum of the three operations: insert, delete, replace
        int insertOp = dfs(word1, word2.substr(1)) + 1;
        int deleteOp = dfs(word1.substr(1), word2) + 1;
        int replaceOp = dfs(word1.substr(1), word2.substr(1)) + 1;

        return memo[word1][word2] = min({insertOp, deleteOp, replaceOp});
    }
};

52128