<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 S1, string S2) {
    int n = S1.size();
    int m = S2.size();

    // Create two arrays to store previous and current row of edit distances
    vector<int> prev(m + 1, 0);
    vector<int> cur(m + 1, 0);

    // Initialize the first row
    for (int j = 0; j <= m; j++) {
        prev[j] = j;
    }

    // Calculate edit distances row by row
    for (int i = 1; i <= n; i++) {
        cur[0] = i; // Initialize the first column of the current row
        for (int j = 1; j <= m; j++) {
            if (S1[i - 1] == S2[j - 1]) {
                // If the characters match, no additional cost
                cur[j] = prev[j - 1];
            } else {
                // Minimum of three choices:
                // 1. Replace the character at S1[i-1] with S2[j-1]
                // 2. Delete the character at S1[i-1]
                // 3. Insert the character at S2[j-1] into S1
                cur[j] = 1 + min(prev[j - 1], min(prev[j], cur[j - 1]));
            }
        }
        // Move current row to previous row for the next iteration
        prev = cur;
    }

    // The value at prev[m] contains the edit distance
    return prev[m];
}

};
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