Updted Edit Distance In Java by adding description
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/**
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Author : SUBHAM SANGHAI
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A Dynamic Programming based solution for Edit Distance problem In Java
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Edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one another,
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by counting the minimum number of operations required to transform one string into the other
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**/
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import java.util.HashMap;
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import java.util.Map;
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/**Description of Edit Distance with an Example:
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Edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one another,
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by counting the minimum number of operations required to transform one string into the other. The
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distance operations are the removal, insertion, or substitution of a character in the string.
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The Distance between "kitten" and "sitting" is 3. A minimal edit script that transforms the former into the latter is:
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kitten → sitten (substitution of "s" for "k")
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sitten → sittin (substitution of "i" for "e")
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sittin → sitting (insertion of "g" at the end).**/
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import java.util.Scanner;
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public class Edit_Distance
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{
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public static int minDistance(String word1, String word2)
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{
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int len1 = word1.length();
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int len2 = word2.length();
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// len1+1, len2+1, because finally return dp[len1][len2]
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int[][] dp = new int[len1 + 1][len2 + 1];
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/* If second string is empty, the only option is to
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insert all characters of first string into second*/
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for (int i = 0; i <= len1; i++)
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{
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dp[i][0] = i;
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}
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/* If first string is empty, the only option is to
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insert all characters of second string into first*/
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for (int j = 0; j <= len2; j++)
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{
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dp[0][j] = j;
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@ -40,6 +55,8 @@
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}
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else
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{
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/* if two characters are different ,
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then take the minimum of the various operations(i.e insertion,removal,substitution)*/
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int replace = dp[i][j] + 1;
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int insert = dp[i][j + 1] + 1;
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int delete = dp[i + 1][j] + 1;
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@ -50,8 +67,11 @@
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}
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}
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}
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/* return the final answer , after traversing through both the strings*/
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return dp[len1][len2];
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}
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// Driver program to test above function
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public static void main(String args[])
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{
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