algorithm/Java
1
0
Fork 0
mirror of https://github.com/TheAlgorithms/Java.git synced 2025-07-22 19:59:09 +08:00
Code Issues Packages Projects Releases Wiki Activity

style: format code (#4212)

Browse Source 
close #4204
This commit is contained in:
acbin
2023-06-09 18:52:05 +08:00
committed by GitHub
parent ad03086f54
commit 00282efd8b
521 changed files with 5233 additions and 7309 deletions
Download Patch File Download Diff File Expand all files Collapse all files

26
src/main/java/com/thealgorithms/dynamicprogramming/PalindromicPartitioning.java
View File

@ -22,9 +22,9 @@ 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
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];
@ -36,7 +36,8 @@ public class PalindromicPartitioning {
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. */
/* 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++) {
@ -48,23 +49,20 @@ public class PalindromicPartitioning {
if (L == 2) {
isPalindrome[i][j] = (word.charAt(i) == word.charAt(j));
} else {
isPalindrome[i][j] = (word.charAt(i) == word.charAt(j)) &&
isPalindrome[i + 1][j - 1];
isPalindrome[i][j]
= (word.charAt(i) == word.charAt(j)) && isPalindrome[i + 1][j - 1];
}
}
}
//We find the minimum for each index
// We find the minimum for each index
for (i = 0; i < len; i++) {
if (isPalindrome[0][i]) {
minCuts[i] = 0;
} else {
minCuts[i] = Integer.MAX_VALUE;
for (j = 0; j < i; j++) {
if (
isPalindrome[j + 1][i] &&
1 + minCuts[j] < minCuts[i]
) {
if (isPalindrome[j + 1][i] && 1 + minCuts[j] < minCuts[i]) {
minCuts[i] = 1 + minCuts[j];
}
}
@ -84,11 +82,7 @@ public class PalindromicPartitioning {
// 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
);
"The minimum cuts needed to partition \"" + word + "\" into palindromes is " + ans);
input.close();
}
}