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Java/DynamicProgramming/EditDistance.java at master · kaishui/Java · GitHub
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EditDistance.java
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EditDistance.java
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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.
*
* <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
();
}
// edit distance problem
public
static
int
editDistance
(
String
s1
,
String
s2
){
int
[][]
storage
=
new
int
[
s1
.
length
() +
1
][
s2
.
length
() +
1
];
return
editDistance
(
s1
,
s2
,
storage
);
}
public
static
int
editDistance
(
String
s1
,
String
s2
,
int
[][]
storage
) {
int
m
=
s1
.
length
();
int
n
=
s2
.
length
();
if
(
storage
[
m
][
n
] >
0
) {
return
storage
[
m
][
n
];
}
if
(
m
==
0
) {
storage
[
m
][
n
] =
n
;
return
storage
[
m
][
n
];
}
if
(
n
==
0
) {
storage
[
m
][
n
] =
m
;
return
storage
[
m
][
n
];
}
if
(
s1
.
charAt
(
0
) ==
s2
.
charAt
(
0
)) {
storage
[
m
][
n
] =
editDistance
(
s1
.
substring
(
1
),
s2
.
substring
(
1
),
storage
);
return
storage
[
m
][
n
];
}
else
{
int
op1
=
editDistance
(
s1
,
s2
.
substring
(
1
),
storage
);
int
op2
=
editDistance
(
s1
.
substring
(
1
),
s2
,
storage
);
int
op3
=
editDistance
(
s1
.
substring
(
1
),
s2
.
substring
(
1
),
storage
);
storage
[
m
][
n
] =
1
+
Math
.
min
(
op1
,
Math
.
min
(
op2
,
op3
));
return
storage
[
m
][
n
];
}
}
}
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