FazBrowse GitHub Viewer
|
Trending
|
URL:
|
Home
Tools:
[Download Repo ZIP]
[View Raw Code]
[Original HTTPS Page]
Java/DynamicProgramming/PalindromicPartitioning.java at master · kaishui/Java · GitHub
kaishui
/
Java
Public
forked from
TheAlgorithms/Java
Notifications
You must be signed in to change notification settings
Fork
0
Star
0
Code
Pull requests
0
Actions
Projects
Security and quality
0
Insights
Additional navigation options
Code
Pull requests
Actions
Projects
Security and quality
Insights
Expand file tree
Breadcrumbs
Java
/
DynamicProgramming
/
PalindromicPartitioning.java
Copy path
More file actions
More file actions
Latest commit
History
History
History
94 lines (82 loc) · 3.71 KB
Breadcrumbs
Java
/
DynamicProgramming
/
PalindromicPartitioning.java
Copy path
File metadata and controls
94 lines (82 loc) · 3.71 KB
Raw
Copy raw file
Download raw file
Open symbols panel
Edit and raw actions
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
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
package
DynamicProgramming
;
import
java
.
util
.
Scanner
;
/**
* @file
* @brief Implements [Palindrome
* Partitioning](https://leetcode.com/problems/palindrome-partitioning-ii/)
* algorithm, giving you the minimum number of partitions you need to make
*
* @details
* palindrome partitioning uses dynamic programming and goes to all the possible
* partitions to find the minimum you are given a string and you need to give
* minimum number of partitions needed to divide it into a number of palindromes
* [Palindrome Partitioning]
* (https://www.geeksforgeeks.org/palindrome-partitioning-dp-17/) overall time
* complexity O(n^2) For example: example 1:- String : "nitik" Output : 2 => "n
* | iti | k" For example: example 2:- String : "ababbbabbababa" Output : 3 =>
* "aba | b | bbabb | ababa"
* @author [Syed] (https://github.com/roeticvampire)
*/
public
class
PalindromicPartitioning
{
public
static
int
minimalpartitions
(
String
word
){
int
len
=
word
.
length
();
/* We Make two arrays to create a bottom-up solution.
minCuts[i] = Minimum number of cuts needed for palindrome partitioning of substring word[0..i]
isPalindrome[i][j] = true if substring str[i..j] is palindrome
Base Condition: C[i] is 0 if P[0][i]= true
*/
int
[]
minCuts
=
new
int
[
len
];
boolean
[][]
isPalindrome
=
new
boolean
[
len
][
len
];
int
i
,
j
,
L
;
// different looping variables
// Every substring of length 1 is a palindrome
for
(
i
=
0
;
i
<
len
;
i
++) {
isPalindrome
[
i
][
i
] =
true
;
}
/* L is substring length. Build the solution in bottom up manner by considering all substrings of length starting from 2 to n. */
for
(
L
=
2
;
L
<=
len
;
L
++) {
// For substring of length L, set different possible starting indexes
for
(
i
=
0
;
i
<
len
-
L
+
1
;
i
++) {
j
=
i
+
L
-
1
;
// Ending index
// If L is 2, then we just need to
// compare two characters. Else need to
// check two corner characters and value
// of P[i+1][j-1]
if
(
L
==
2
)
isPalindrome
[
i
][
j
] = (
word
.
charAt
(
i
) ==
word
.
charAt
(
j
));
else
{
if
((
word
.
charAt
(
i
) ==
word
.
charAt
(
j
)) &&
isPalindrome
[
i
+
1
][
j
-
1
])
isPalindrome
[
i
][
j
] =
true
;
else
isPalindrome
[
i
][
j
]=
false
;
}
}
}
//We find the minimum for each index
for
(
i
=
0
;
i
<
len
;
i
++) {
if
(
isPalindrome
[
0
][
i
] ==
true
)
minCuts
[
i
] =
0
;
else
{
minCuts
[
i
] =
Integer
.
MAX_VALUE
;
for
(
j
=
0
;
j
<
i
;
j
++) {
if
(
isPalindrome
[
j
+
1
][
i
] ==
true
&&
1
+
minCuts
[
j
] <
minCuts
[
i
])
minCuts
[
i
] =
1
+
minCuts
[
j
];
}
}
}
// Return the min cut value for complete
// string. i.e., str[0..n-1]
return
minCuts
[
len
-
1
];
}
public
static
void
main
(
String
[]
args
) {
Scanner
input
=
new
Scanner
(
System
.
in
);
String
word
;
System
.
out
.
println
(
"Enter the First String"
);
word
=
input
.
nextLine
();
// ans stores the final minimal cut count needed for partitioning
int
ans
=
minimalpartitions
(
word
);
System
.
out
.
println
(
"The minimum cuts needed to partition
\"
"
+
word
+
"
\"
into palindromes is "
+
ans
);
input
.
close
();
}
}
Back
|
FazBrowse Home
|
New Git URL