<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) {
        int** cache = new int*[word1.size()+1];
        for (int i = 0; i < word1.size()+1; i++)
            cache[i] = new int[word2.size()+1];
        
        for (int i = 0; i < word1.size()+1; i++) {
            for (int j = 0; j < word2.size()+1; j++) {
                cache[i][j] = INT_MAX;
            }
        }

        for (int i = 0; i < word1.size()+1; i++) {
            cache[i][word2.size()] = word1.size() - i;
        }

        for (int j = 0; j < word2.size()+1; j++) {
            cache[word1.size()][j] = word2.size() - j;
        }

        for (int i = word1.size()-1; i >= 0; i--) {
            for (int j = word2.size()-1; j >= 0; j--) {
                if (word1[i] == word2[j]) {
                    cache[i][j] = cache[i+1][j+1];
                } else {
                    cache[i][j] = std::min({cache[i+1][j], cache[i+1][j+1], cache[i][j+1]}) + 1;
                }
            }
        }

        return cache[0][0];
    }
};
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