Understand Palindrome Partitioning Problem

Problem Name: Palindrome Partitioning
Problem Description:

Palindrome Partitioning Problem

Problem Statement:

Given a string s, the task is to partition the string into all possible palindrome substrings. We need to find and return all possible valid partitions where each substring is a palindrome.

Approach:

1. ispalindrome Function:

The ispalindrome function checks whether a given substring of string s is a palindrome.

Parameters:

  • s: The input string.
  • start: The starting index of the substring.
  • end: The ending index of the substring.

Implementation:

  • This function uses a two-pointer approach to compare characters from both ends of the substring.
  • If any characters don’t match, it returns false, indicating the substring is not a palindrome.
  • If all characters match, it returns true and prints the palindrome substring.

2. calculate Function:

The calculate function recursively generates all valid partitions of the input string s.

Parameters:

  • index: The current index in the string.
  • t: A vector of vectors that stores the valid palindrome partitions.
  • s: The input string.
  • path: A vector that stores the current partition path.

Implementation:

  • The function works recursively and explores all possible substrings starting from the index.
  • If the substring from index to i is a palindrome, we add it to the path.
  • The function then recursively calls itself with the updated index and the current path.
  • If the index reaches the end of the string, we’ve found a valid partition, and we add it to the result (t).
  • Backtracking is performed by removing the last added palindrome from the path.

3. partition Function:

This is the main function that initializes the process of palindrome partitioning.

Parameter:

  • s: The input string.

Implementation:

  • The function initializes an empty vector of vectors (t) to store the valid palindrome partitions.
  • It also initializes an empty vector (path) to store the current partition path.
  • The calculate function is called with initial parameters to start the recursive partition generation process.
  • Finally, the function returns the vector t containing all valid palindrome partitions.

Test Cases

Test Case 1: Basic Case with Single Palindrome

Input:
s = "aab"

Output:
[["a", "a", "b"], ["aa", "b"]]

Explanation:
The string "aab" can be partitioned in two valid ways:

  1. "a", "a", "b"
  2. "aa", "b"

Test Case 2: String with All Palindromes

Input:
s = "a"

Output:
[["a"]]

Explanation:
The string "a" is already a palindrome, so the only partition is itself.

Test Case 3: String with Multiple Palindromes

Input:
s = "racecar"

Output:
[["r", "a", "c", "e", "c", "a", "r"], ["racecar"]]

Explanation:
The string "racecar" can be partitioned in two valid ways:

  1. "r", "a", "c", "e", "c", "a", "r"
  2. "racecar" (the whole string is a palindrome)

Test Case 4: No Palindrome Substring

Input:
s = "abc"

Output:
[["a", "b", "c"]]

Explanation:
Since no two characters form a valid palindrome substring except for the single characters, the only valid partition is each character individually.

Test Case 5: Larger Input with Palindromes

Input:
s = "aabb"

Output:
[["a", "a", "b", "b"], ["aa", "b", "b"], ["a", "a", "bb"]]

Explanation:
The string "aabb" can be partitioned into the following valid palindrome partitions:

  1. "a", "a", "b", "b"
  2. "aa", "b", "b"
  3. "a", "a", "bb"
Category:
  • Strings
  • DP
  • Backtracking
Programming Language:
  • Java
Reference Link:

https://leetcode.com/problems/palindrome-partitioning/description/

Online IDE

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Java
Output:
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Tracking code (View only. In case you want to track the code, click this button):
Main Function:
INPUT: s = "aab";
OUTPUT: [ [ "a", "a" , "b" ],[ "aa","b"] ]
public static void main(String args[]) {
 String s = "aabb";
 List < List < String >> ans = partition(s);
int n = ans.size();
System.out.println("The Palindromic partitions are :-");
 System.out.print(" [ ");
for (int i = 0; i < ans.size(); i++) {
System.out.print("[ ");
 for (int j = 0; j < ans.get(i).size(); j++) {
System.out.print(ans.get(i).get(j) + " ");
}//Loop End
 System.out.print("] ");
}//Loop End
System.out.print("]");
 }//function end
Helper Function:
public static List < List < String >> partition(String s) {
 List < List < String >> res = new ArrayList < > ();
 List < String > path = new ArrayList < > ();
calculate(0, s, path, res);
return res;
}//function end
Utility Functions and Global variables:
    static void calculate(int index, String s, List < String > path, List < List < String >> res) {
 if (index == s.length()) {
res.add(new ArrayList < > (path));
return;
}//If End
for (int i = index; i < s.length(); ++i) {
 if (isPalindrome(s, index, i)) {
path.add(s.substring(index, i + 1));
calculate(i + 1, s, path, res);
path.remove(path.size() - 1);
}//If End
}//Loop End
}//function end
 static boolean isPalindrome(String s, int start, int end) {
while (start <= end) {
if (s.charAt(start++) != s.charAt(end--)){
return false;
}//If End
}//Loop End
return true;
}//function end
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