<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 dp[501][501];
    int solve1(int i,int j,string s1,string s2)
    {
        if(i >= s1.size()) {
            return (int)s2.size() - j;
        }
        if(j >= s2.size()) {
            return (int)s1.size() - i;
        }
        if(s1[i] == s2[j]) {
            if(dp[i+1][j+1] != -1) {
                return dp[i+1][j+1];
            }
            return dp[i+1][j+1] = solve1(i+1,j+1,s1,s2);
        }
        else {
           if(dp[i+1][j+1] == -1) { 
            dp[i+1][j+1] = solve1(i+1,j+1,s1,s2); // replace on ith
           } 
           if(dp[i][j+1] == -1) {
            dp[i][j+1] = solve1(i,j+1,s1,s2); // add ith
           }
           if(dp[i+1][j] == -1) {
            dp[i+1][j] = solve1(i+1,j,s1,s2); // delete ith
           }
           return 1 + min({dp[i+1][j+1],dp[i][j+1],dp[i+1][j]});
        }
    }
    int minDistance(string word1, string word2) {
      memset(dp, -1, sizeof(dp));
      int ans = solve1(0,0,word1,word2);
      return ans;
    }
};
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