Palindrome Partitioning Using Backtracking in DSA Java - Build from Scratch

Palindrome Partitioning Using Backtracking

📖 Scenario: Imagine you have a word, and you want to split it into parts where each part reads the same forwards and backwards. This is called palindrome partitioning. For example, the word "noon" can be split as "n" | "oo" | "n" where each part is a palindrome.We will write a program to find all such ways to split a word using a method called backtracking, which tries all possible splits and keeps only the good ones.

🎯 Goal: Build a Java program that takes a word and finds all possible palindrome partitions using backtracking. You will create the data, set up helper variables, write the backtracking logic, and finally print all palindrome partitions.

📋 What You'll Learn

Create a string variable with the exact value "aab"

Create a list to hold the final list of palindrome partitions

Write a helper function to check if a substring is a palindrome

Write a backtracking function to find all palindrome partitions

Print all palindrome partitions in the format [["a", "a", "b"], ["aa", "b"]]

💡 Why This Matters

🌍 Real World

Palindrome partitioning is useful in text processing, DNA sequence analysis, and solving puzzles where symmetrical patterns matter.

💼 Career

Understanding backtracking and palindrome checks helps in coding interviews and problems involving string manipulation and recursion.

1

Create the input string

Create a string variable called s and set it to the exact value "aab".

DSA Java

// Your code here

Hint

Use String s = "aab"; to create the input string.

2

Create the result list to hold partitions

Add a variable called result of type List> initialized as a new ArrayList<>() to hold all palindrome partitions.

DSA Java

public class PalindromePartitioning {
    public static void main(String[] args) {
        String s = "aab";
        // Your code here
    }
}

Hint

Use List> result = new ArrayList<>(); to create the list.

3

Write the backtracking and palindrome check methods

Add two methods inside the class: isPalindrome(String s, int start, int end) that returns true if the substring is a palindrome, and backtrack(String s, int start, List current, List> result) that finds all palindrome partitions using backtracking. Call backtrack(s, 0, new ArrayList<>(), result) inside main.

DSA Java

import java.util.ArrayList;
import java.util.List;

public class PalindromePartitioning {
    public static void main(String[] args) {
        String s = "aab";
        List> result = new ArrayList();
        // Your code here
    }

    // Add isPalindrome method here

    // Add backtrack method here
}

Hint

Use a loop from start to s.length() and check palindrome with isPalindrome. Add substring to current, recurse, then remove last added.

4

Print all palindrome partitions

Add a print statement inside main to display the result list after backtracking finishes.

DSA Java

import java.util.ArrayList;
import java.util.List;

public class PalindromePartitioning {
    public static void main(String[] args) {
        String s = "aab";
        List> result = new ArrayList();
        backtrack(s, 0, new ArrayList(), result);
        // Your code here
    }

    public static boolean isPalindrome(String s, int start, int end) {
        while (start  current, List> result) {
        if (start == s.length()) {
            result.add(new ArrayList(current));
            return;
        }
        for (int end = start; end < s.length(); end++) {
            if (isPalindrome(s, start, end)) {
                current.add(s.substring(start, end + 1));
                backtrack(s, end + 1, current, result);
                current.remove(current.size() - 1);
            }
        }
    }

Hint

Use System.out.println(result); to print the list of palindrome partitions.

Practice

1. What is the main goal of palindrome partitioning using backtracking?

easy

A. To count the number of palindromes in a string

B. To find the longest palindrome substring in a string

C. To split a string into all possible palindrome substrings

D. To reverse the string and check if it is a palindrome

Solution

1. Step 1: Understand palindrome partitioning

   Palindrome partitioning means dividing a string into parts where each part is a palindrome.

2. Step 2: Role of backtracking

   Backtracking helps explore all possible ways to split the string into palindrome parts.

3. Final Answer:

   To split a string into all possible palindrome substrings -> Option C

4. Quick Check:

   Palindrome partitioning = splitting into palindrome substrings [OK]

Hint: Backtracking explores all palindrome splits [OK]

Common Mistakes:

• Confusing palindrome partitioning with longest palindrome substring
• Thinking it only counts palindromes instead of listing partitions
• Assuming it reverses the whole string

2. Which of the following is the correct Java method signature for a backtracking function that partitions a string into palindromes?

easy

A. boolean backtrack(String s, int start, List<String> current)

B. void backtrack(String s, int start, List result, List<String> current)

C. int backtrack(String s, int start, List<String> result)

D. String backtrack(String s, int start, List result)

Solution

1. Step 1: Analyze method parameters

   The method needs the string, current start index, a list to store all partitions, and the current partition path.

2. Step 2: Check return type

   Backtracking methods usually return void because they build results via parameters.

3. Final Answer:

   void backtrack(String s, int start, List result, List<String> current) -> Option B

4. Quick Check:

   Backtracking method returns void and uses lists [OK]

Hint: Backtracking returns void and uses lists for results [OK]

Common Mistakes:

• Using wrong return type like int or boolean
• Missing current partition list parameter
• Returning String instead of void

3. What will be the output partitions for the input string "aab" using palindrome partitioning?

medium

A. [["aab"]]

B. [["aa", "ab"]]

C. [["a", "ab"]]

D. [["a", "a", "b"], ["aa", "b"]]

Solution

1. Step 1: Identify palindrome substrings

   "a" and "aa" are palindromes; "b" is a palindrome; "ab" is not.

2. Step 2: List all palindrome partitions

   Partitions are ["a", "a", "b"] and ["aa", "b"].

3. Final Answer:

   [["a", "a", "b"], ["aa", "b"]] -> Option D

4. Quick Check:

   Only palindrome splits included [OK]

Hint: Check each substring palindrome before partitioning [OK]

Common Mistakes:

• Including non-palindrome substrings like "ab"
• Missing the partition with "aa"
• Returning only one partition

4. Identify the error in this Java backtracking snippet for palindrome partitioning:

void backtrack(String s, int start, List result, List<String> current) {
  if (start == s.length()) {
    result.add(current);
    return;
  }
  for (int end = start; end < s.length(); end++) {
    if (isPalindrome(s, start, end)) {
      current.add(s.substring(start, end + 1));
      backtrack(s, end + 1, result, current);
      current.remove(current.size());
    }
  }
}

medium

A. Incorrect removal index in current.remove()

B. Missing base case for empty string

C. Wrong substring indices in s.substring()

D. Backtracking call uses wrong start index

Solution

1. Step 1: Check removal from current list

   List indices start at 0, so removing at current.size() causes IndexOutOfBoundsException.

2. Step 2: Correct removal index

   Should remove last element with current.remove(current.size() - 1).

3. Final Answer:

   Incorrect removal index in current.remove() -> Option A

4. Quick Check:

   Remove last element with size()-1 [OK]

Hint: Remove last added element with size()-1 index [OK]

Common Mistakes:

• Removing element at size() instead of size()-1
• Confusing substring indices
• Incorrect recursive call start index

5. Given the string "racecar", how many palindrome partitions will the backtracking algorithm generate?

hard

A. 6

B. 8

C. 4

D. 10

Solution

1. Step 1: Identify palindrome substrings in "racecar"

   Palindromes include "r", "a", "c", "e", "cec", "aceca", "racecar" and their combinations.

2. Step 2: Count all palindrome partitions

   Partitions are:
   ["racecar"],
   ["r", "aceca", "r"],
   ["r", "a", "cec", "a", "r"],
   ["r", "a", "c", "e", "c", "a", "r"],
   and others including splits like ["r", "a", "c", "ec", "a", "r"] are invalid because "ec" is not palindrome.
   Counting all valid partitions, the total is 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Count all palindrome partitions = 6 [OK]

Hint: Count all palindrome splits including nested palindromes [OK]

Common Mistakes:

• Counting only single-character partitions
• Missing longer palindrome substrings
• Double counting partitions