<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 dp[505][505];
    
    int f(string &s1, string &s2, int i, int j) {
        // Base cases
        if (i == s1.size()) return s2.size() - j; // Remaining insertions to match s2
        if (j == s2.size()) return s1.size() - i; // Remaining deletions to match s1
        
        // Memoization check
        if (dp[i][j] != -1) return dp[i][j];
        
        // If characters match, move both indices forward
        if (s1[i] == s2[j]) {
            return dp[i][j] = f(s1, s2, i+1, j+1);
        }
        
        // Calculate the minimum cost considering all operations: 
        // substitution, deletion, and insertion
        return dp[i][j] = min({
            1 + f(s1, s2, i+1, j+1), // substitution
            1 + f(s1, s2, i+1, j),   // deletion
            1 + f(s1, s2, i, j+1)    // insertion
        });
    }
    
    int minDistance(string word1, string word2) {
        memset(dp, -1, sizeof(dp)); // Initialize the dp array
        return f(word1, word2, 0, 0);
    }
};

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