/**
    Author : SUBHAM SANGHAI
    A Dynamic Programming based solution for Edit Distance problem In Java
    **/

    /**Description of Edit Distance with an Example:

    Edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one another, 
    by counting the minimum number of operations required to transform one string into the other. The
    distance operations are the removal, insertion, or substitution of a character in the string.


    The Distance between "kitten" and "sitting" is 3. A minimal edit script that transforms the former into the latter is:

    kitten → sitten (substitution of "s" for "k")
    sitten → sittin (substitution of "i" for "e")
    sittin → sitting (insertion of "g" at the end).**/

    import java.util.Scanner;
    public class Edit_Distance
    {

      

      public static int minDistance(String word1, String word2) 
      {
    	int len1 = word1.length();
    	int len2 = word2.length();
     	// len1+1, len2+1, because finally return dp[len1][len2]
    	int[][] dp = new int[len1 + 1][len2 + 1];
     	/* If second string is empty, the only option is to
   	  insert all characters of first string into second*/
    	for (int i = 0; i <= len1; i++) 
    	{
    		dp[i][0] = i;
    	}
     	/* If first string is empty, the only option is to
   	  insert all characters of second string into first*/
    	for (int j = 0; j <= len2; j++) 
    	{
    		dp[0][j] = j;
    	}
     	//iterate though, and check last char
    	for (int i = 0; i < len1; i++) 
    	{
    		char c1 = word1.charAt(i);
    		for (int j = 0; j < len2; j++) 
    		{
    			char c2 = word2.charAt(j);
     			//if last two chars equal
    			if (c1 == c2)
    			{
    				//update dp value for +1 length
    				dp[i + 1][j + 1] = dp[i][j];
    			} 
    			else 
    			{
			/* if two characters are different ,
			then take the minimum of the various operations(i.e insertion,removal,substitution)*/
    				int replace = dp[i][j] + 1;
    				int insert = dp[i][j + 1] + 1;
    				int delete = dp[i + 1][j] + 1;
     
    				int min = replace > insert ? insert : replace;
    				min = delete > min ? min : delete;
    				dp[i + 1][j + 1] = min;
    			}
    		}
    	}
	/* return the final answer , after traversing through both the strings*/
    	return dp[len1][len2];
      }


    	// Driver program to test above function
    	public static void main(String args[])
    	{
    		Scanner input = new Scanner(System.in);
    		String s1,s2;
    		System.out.println("Enter the First String");
    		s1 = input.nextLine();
    		System.out.println("Enter the Second String");
    		s2 = input.nextLine();
    		//ans stores the final Edit Distance between the two strings
    		int ans=0;
    		ans=minDistance(s1,s2);
    		System.out.println("The minimum Edit Distance between \"" + s1 + "\" and \"" + s2 +"\" is "+ans);
    	}
    }