<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 n = word1.size() ; 
        int m  = word2.size() ; 
        if(m == 0 && n ==0   )return  0 ;  
        if(m ==  0 || n == 0 )return 1; 
        vector<vector<int>>dp(n  + 5  , vector<int>(m + 5)) ; 
        for(int i = 1 ; i <= m  ; i++)dp[0][i] = i ; 
        for(int i = 1 ; i <= n ; i++)dp[i][0] =  i ; 
        dp[0][0] =  0 ; 
        for(int i = 1   ; i <= n  ; i++){
            for(int j = 1  ;  j  <=  m ; j++){
                  if(word1[i-1] == word2[j-1] ){
                      dp[i][j] = dp[i - 1][j - 1 ] ; 
                  }
                  else{
                       int top = dp[i - 1][j] ; 
                       int back =  dp[i][j - 1] ; 
                       int tl = dp[i - 1][j - 1]; 
                       dp[i][j]= min({top ,  back , tl}) + 1 ; 
                  }
            }
        }
        return dp[n][m];
    }
};
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