<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:
    string s1,s2;
    map<pair<int,int>,int> mp;
    int m(int i, int j){
        return s1[i]!=s2[j];
    }
    int cost(/*int i, int j*/){
       /* if (i==-1 && j==-1)return 0;
        if (i==-1)return j+1;
        if (j==-1)return i+1;
        if (mp.count({i,j}))return mp[{i,j}];
        return mp[{i,j}] = min(cost(i,j-1)+1,min(cost(i-1,j)+1,cost(i-1,j-1)+m(i,j)));*/
        mp[{-1,-1}] = 0;
        for (int j = 0; j<s2.size(); ++j){
            mp[{-1,j}] = j+1;
        }
        for (int i = 0; i<s1.size(); ++i){
            mp[{i,-1}] = i+1;
        }
        for (int i = 0; i<s1.size(); ++i){
            for (int j = 0; j<s2.size(); ++j){
                mp[{i,j}] = min(mp[{i,j-1}]+1,min(mp[{i-1,j}]+1,mp[{i-1,j-1}]+m(i,j)));
            }
        }
        return mp[{s1.size()-1,s2.size()-1}];
    }
    /*void cost_print(int i, int j){
        if (i==-1 && j==-1)return;
        if (i==-1){
            cout<<"add remaining of j"<<endl;
            return;
        }
        if (j==-1){
            cout<<"delete remaining of i"<<endl;
            return;
        }
        if (s1[i]==s2[j]){
            cost_print(i-1,j-1);
            return;
        }
        int optimal = cost(i,j);
        if (optimal == (cost(i,j-1)+1)){
            cout<<"insert at "<<i<<endl;
            cost_print(i,j-1);
        }else if (optimal == (cost(i-1,j)+1)){
            cout<<"delete at "<<i<<endl;
            cost_print(i-1,j);
        }else {
            cout<<"replace at "<<i<<endl;
            cost_print(i-1,j-1);
        }
    }*/
    
    int minDistance(string word1, string word2) {
        s1 = word1;
        s2 = word2;
        int lol = cost(/*word1.size()-1,word2.size()-1*/);
        //cost_print(s1.size()-1,s2.size()-1);
        return lol;
    }
};
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