119 lines
3.8 KiB
Java
119 lines
3.8 KiB
Java
package DynamicProgramming;
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/**
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* A DynamicProgramming based solution for Edit Distance problem In Java Description of Edit
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* Distance with an Example:
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*
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* <p>Edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one
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* another, by counting the minimum number of operations required to transform one string into the
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* other. The distance operations are the removal, insertion, or substitution of a character in the
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* string.
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*
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* <p>
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*
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* <p>The Distance between "kitten" and "sitting" is 3. A minimal edit script that transforms the
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* former into the latter is:
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*
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* <p>kitten → sitten (substitution of "s" for "k") sitten → sittin (substitution of "i" for "e")
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* sittin → sitting (insertion of "g" at the end).
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*
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* @author SUBHAM SANGHAI
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*/
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import java.util.Scanner;
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public class EditDistance {
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public static int minDistance(String word1, String word2) {
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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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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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dp[0][j] = j;
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}
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// iterate though, and check last char
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for (int i = 0; i < len1; i++) {
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char c1 = word1.charAt(i);
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for (int j = 0; j < len2; j++) {
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char c2 = word2.charAt(j);
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// if last two chars equal
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if (c1 == c2) {
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// update dp value for +1 length
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dp[i + 1][j + 1] = dp[i][j];
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} else {
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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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int min = replace > insert ? insert : replace;
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min = delete > min ? min : delete;
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dp[i + 1][j + 1] = min;
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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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public static void main(String[] args) {
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Scanner input = new Scanner(System.in);
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String s1, s2;
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System.out.println("Enter the First String");
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s1 = input.nextLine();
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System.out.println("Enter the Second String");
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s2 = input.nextLine();
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// ans stores the final Edit Distance between the two strings
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int ans = minDistance(s1, s2);
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System.out.println(
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"The minimum Edit Distance between \"" + s1 + "\" and \"" + s2 + "\" is " + ans);
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input.close();
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}
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// edit distance problem
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public static int editDistance(String s1, String s2){
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int[][] storage = new int[s1.length() + 1][s2.length() + 1];
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return editDistance(s1, s2,storage);
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}
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public static int editDistance(String s1, String s2, int[][] storage) {
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int m = s1.length();
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int n = s2.length();
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if (storage[m][n] > 0) {
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return storage[m][n];
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}
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if (m== 0) {
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storage[m][n] = n;
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return storage[m][n];
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}
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if (n== 0) {
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storage[m][n] = m;
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return storage[m][n];
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}
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if (s1.charAt(0) == s2.charAt(0)) {
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storage[m][n] = editDistance(s1.substring(1), s2.substring(1), storage);
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return storage[m][n];
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} else {
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int op1 = editDistance(s1, s2.substring(1),storage);
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int op2 = editDistance(s1.substring(1), s2,storage);
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int op3 = editDistance(s1.substring(1),s2.substring(1),storage);
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storage[m][n] = 1 + Math.min(op1, Math.min(op2, op3));
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return storage[m][n];
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}
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}
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}
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