forked from TheAlgorithms/Java
-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathEditDistance.java
More file actions
79 lines (72 loc) · 2.97 KB
/
EditDistance.java
File metadata and controls
79 lines (72 loc) · 2.97 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
package DynamicProgramming;
/**
* A DynamicProgramming based solution for Edit Distance problem In Java
* Description of Edit Distance with an Example:
* <p>
* 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.
* <
73D4
;p>
* <p>
* The Distance between "kitten" and "sitting" is 3. A minimal edit script that transforms the former into the latter is:
* <p>
* kitten → sitten (substitution of "s" for "k")
* sitten → sittin (substitution of "i" for "e")
* sittin → sitting (insertion of "g" at the end).
*
* @author SUBHAM SANGHAI
**/
import java.util.Scanner;
public class EditDistance {
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];
}
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 = minDistance(s1, s2);
System.out.println("The minimum Edit Distance between \"" + s1 + "\" and \"" + s2 + "\" is " + ans);
input.close();
}
}