<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 len1 = word1.size();
       int len2 = word2.size();
       unordered_map<int,map<int,int>>dp;
       for(int i = 0; i <= len1; i++) dp[i][0] = i;
       for(int j = 0; j <= len2; j++) dp[0][j] = j;

       for(int i = 1; i <= len1; i++){
         for(int j = 1; j <= len2; j++){
            int flag = (word1[i-1] == word2[j-1]) ? 0 : 1;
            dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + 1;
            dp[i][j] = min(dp[i][j], dp[i-1][j-1] + flag);
            //cout<<"index ("<<i<<" , "<<j<<") = "<<dp[i][j]<<endl;
         }
       }

       return dp[len1][len2];
    }
};
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