<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>

int getMin(vector<int> vals) {
  int min = vals[0];

  for (int i=1; i<vals.size(); i++) {
    if (vals[i] < min) {
      min = vals[i];
    }
  }

  return min;
}

class Solution {
public:
    int minDistance(string word1, string word2) {
      int m = word1.size();
      int n = word2.size();

      vector<vector<int>> dp;

      for (int i=0; i<m+1; i++) {
        vector<int> row(n+1);
        dp.push_back(row);
      }    

      int dxVals[3] = {-1, 0, -1};
      int dyVals[3] = {0, -1, -1};

      for (int i=0; i<=m; i++) {
        for (int j=0; j<=n; j++) {
          vector<int> vals;  

          for (int k=0; k<3; k++) {
            int dx = dxVals[k];
            int dy = dyVals[k];

            if ((i+dx >= 0) && (j+dy >= 0)) {
              if (dx==0 || dy==0) {
                vals.push_back(1+dp[i+dx][j+dy]);   
              } else {
                int equalBit = (word1[m-i] != word2[n-j]) ? 1 : 0;
                vals.push_back(equalBit+dp[i-1][j-1]);
              }
            }
          }

          if (i==0 && j==0) {
            vals.push_back(0);
          }

          dp[i][j] = getMin(vals);
        }
      }

      return dp[m][n];
    }
};
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