<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:
long long dp[1005][1005];
    Solution() {
        memset(dp, -1, sizeof(dp));
    }
    int minDistance(string word1, string word2) {
        int m = word1.length();
        int n = word2.length();
        return solve(word1,word2, m, n);
    }
    int solve(string& a, string& b, int m, int n){
        if( m == 0) return n;
        if(n == 0) return m;
        if(dp[m][n]!=-1) return dp[m][n];
        if(a[m-1] == b[n-1]){
            return dp[m][n] = solve(a, b, m-1, n-1);
        }
        int op1 = solve(a,b, m,n-1);
        int op2 = solve(a,b,m-1,n-1);
        int op3 = solve(a,b,m-1,n);
        return dp[m][n]= 1 + min({op1,op2,op3});
    }
};
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