<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 solve(int i, int j, string &s, string &t, vector<vector<double>> &dp){
        if(i==0)  return j;//  as there are j+1 operations required to convert str1 to str2
        if(j==0) return i;// as str2 is exausted, i still have i+1 string still to be deleted
        
        if(dp[i][j]!=-1) return dp[i][j];
        //match
        if(s[i-1]==t[j-1]){
            return dp[i][j]=0+solve(i-1, j-1, s,t, dp);
        }
        else{
           int insertion = 1 + solve(i, j-1, s,t, dp);
           int deletion = 1 + solve(i-1, j, s, t, dp);
           int replace = 1 + solve(i-1,j-1, s, t, dp);
           return dp[i][j]=min(insertion, min(deletion, replace));
        }
    }
    int tabulation(string &s, string &t){
        int n=s.size();
        int m=t.size();
        vector<vector<double>> dp(n+1, vector<double>(m+1, 0));

        //base condition
       // if(i==0) return j;
       for(int j=0;j<=m;j++){
            dp[0][j]=j;
       }
       //if(j==0) return i;
       for(int i=0;i<=n;i++){
            dp[i][0]=i;
       }
       for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
             if(s[i-1]==t[j-1]){
                dp[i][j]=dp[i-1][j-1];
            }
            else{
                int insertion = 1 + dp[i][j-1];
                int deletion = 1 + dp[i-1][j];
                int replace = 1 + dp[i-1][j-1];
                dp[i][j]=min(insertion, min(deletion, replace));
            }
        }
       }
       return dp[n][m];
    }
    int minDistance(string s, string t) {
        int n=s.size();
        int m=t.size();
        // vector<vector<double>> dp(n+1, vector<double>(m+1, -1));
        // return solve(n, m, s,t, dp);
        return tabulation(s,t);
    }
};
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