palindrome

package palindrome;

/*Problem Statement


A palindrome is a string that reads the same forward and backward.

A palindrome substring is a contiguous sequence of characters taken from a string that form a palindrome.

A palindrome substring of a string is maximal if we can't extend it to get a bigger palindrome substring.

For example, string "acdadc" has 2 maximal palindrome substrings - "a" (the first one) and "cdadc".

All other palindrome substrings (like "dad" or "c") can be extended to "cdadc", so they are not maximal.

You will be given a String[] str.

Concatenate the elements of str to form one long string, and return the number of maximal palindrome substrings contained in that string.

Definition

Class: MaximalPalindromeSubstrings
Method: countMaximalPalindromeSubstrings
Parameters: String[]
Returns: int
Method signature: int countMaximalPalindromeSubstrings(String[] str)

(be sure your method is public)

Constraints
- str will contain between 1 and 50 elements, inclusive.
- Each element of str will contain between 1 and 50 characters, inclusive.
- Each character of each element of str will be a lowercase letter ('a'-'z').
Examples
0) {"acdadc"} Returns: 2 The example from the problem statement.
1) {"ababab"} Returns: 2 The two maximal palindrome substrings here are "ababa" and "babab".
2) {"aaaa","bbb","axxx"} Returns: 3 Remember to use the whole input!
3) {"abacabbacaacdbdcacxcbbbbcabababacccbazhhaahh"} Returns: 14

This problem statement is the exclusive and proprietary property of TopCoder, Inc.
Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited.
(c)2003, TopCoder, Inc. All rights reserved. */

public class MaximalPalindromeSubstrings {


public static int countMaximalPalindromeSubstrings(String[] str) {

int length = 0;
for(int i = 0; i < str.length; i++) {
length = length + str[i].length();
}

char[] array = new char[length];

int p = 0;
for(int i = 0; i < str.length; i++) {
char[] tempArray = str[i].toCharArray();
for(int j = 0; j < tempArray.length; j++) {
array[p] = tempArray[j];
p++;
}
}

p = 0;
while(p<length) {
int index = 0;
while(p - index >=0 && p + index < length) {
if(array[p - index] == array[p + index]) index++;
else break;
}
index--;
save(p - index, p + index);

index = 0;
if((p + 1 < length) && (array[p] == array[p + 1])) {
while(p - index >=0 && p + 1 + index < length) {
if(array[p - index] == array[p + 1 + index]) index++;
else break;
}
index--;
save(p - index, p + 1 + index);
}
p++;
}

print(array);
return root -1;
}


// Stack for the result storage.
private static Pair[] result = new Pair[100];

private static int root = 0;

static class Pair {
int start = 0;
int end = 0;

Pair(int i, int j) {
start = i;
end = j;
}
}

/*
* Check if there is a larger ranger or smaller range.
* Erase the smaller range if needed.
*/
private static void save(int i, int j) {
if(root == 0) {
Pair pair = new Pair(i, j);
result[0] = pair;
root++;
}
else {
int cursor = root - 1;

if (result[cursor].start <= i && result[cursor].end >= j) return;

while(cursor >= 0) {
if(result[cursor].start >= i && result[cursor].end <= j) {
cursor--;
result[root - 1] = null;
root--;
}
else break;
}

Pair pair = new Pair(i, j);
result[root] = pair;
root++;

}
}

/*
* print the result out.
*/
private static void print(char[] array) {
for(int i = 0; i < root; i++) {
for(int j = result[i].start; j <= result[i].end; j++) {
System.out.print(array[j]);
}
System.out.print("\n");
}
}
}




public class Main {

/**
* @param args
*/
public static void main(String[] args) {

try {

String[] a = {"aaaa","bbb","axxx"};
String b = "abacabbacaacdbdcacxcbbbbcabababacccbazhhaahh";
MaximalPalindromeSubstrings.countMaximalPalindromeSubstrings(new String[]{b});
} catch(Exception e) {
e.printStackTrace();
}

}
}

### XTUOJ Perfect Palindrome Problem Analysis For the **Perfect Palindrome** problem on the XTUOJ platform, understanding palindromes and string manipulation algorithms plays a crucial role. A palindrome refers to a word, phrase, number, or other sequences of characters which reads the same backward as forward[^1]. The challenge typically involves checking whether a given string meets specific conditions to be considered a perfect palindrome. In many similar problems, preprocessing steps such as converting all letters into lowercase (or uppercase) can simplify subsequent checks by ensuring case insensitivity during comparison operations. Additionally, removing non-alphanumeric characters ensures that only relevant symbols participate in determining if the sequence forms a valid palindrome[^2]. To determine if a string is a perfect palindrome, one approach iterates from both ends towards the center while comparing corresponding elements until reaching the midpoint without encountering mismatches: ```python def is_perfect_palindrome(s): cleaned_string = ''.join(char.lower() for char in s if char.isalnum()) left_index = 0 right_index = len(cleaned_string) - 1 while left_index < right_index: if cleaned_string[left_index] != cleaned_string[right_index]: return False left_index += 1 right_index -= 1 return True ``` This function first creates `cleaned_string`, stripping away any irrelevant characters and normalizing cases. Then through iteration with two pointers moving inward simultaneously (`left_index` starting at position 0 and `right_index` initially set to the last index), comparisons occur between pairs of opposing positions within the processed input string. If every pair matches perfectly throughout this process, then the original string qualifies as a "perfect palindrome".
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