Implementation of Edit Distance in Java

Dynamic Programming/Edit_Distance.java

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+    /**
+    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);
+    	}
+    }