Palindrome Partitioning Using Backtracking in DSA Python - 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"] or ["noon"].

🎯 Goal: You will build a program that finds all the ways to split a given word into palindrome parts using backtracking. This helps understand how to explore all possible splits and check for palindromes.

📋 What You'll Learn

Create a string variable with a specific word

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

Use backtracking to find all palindrome partitions

Print the list of all palindrome partitions

💡 Why This Matters

🌍 Real World

Palindrome partitioning helps in text processing, DNA sequence analysis, and finding symmetrical patterns in data.

💼 Career

Understanding backtracking and palindrome checks is useful for coding interviews and solving complex string problems in software development.

1

Create the input string

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

DSA Python

# Create the string variable 'word' with value "aab"
# Your code here

Hint

Use word = "aab" to create the string.

2

Create a palindrome check function

Create a function called is_palindrome that takes a string s and returns True if s is the same forwards and backwards, otherwise False. Use slicing to check this.

DSA Python

word = "aab"
# Define the function 'is_palindrome(s)' below
# Your code here

Hint

Use return s == s[::-1] to check palindrome.

3

Implement backtracking to find palindrome partitions

Create a function called backtrack that takes start index and path list. Use a for loop with end from start + 1 to len(word) + 1. Inside the loop, get substring word[start:end]. If it is a palindrome (use is_palindrome), append it to path, call backtrack recursively with end and updated path, then remove the last element from path to backtrack. Create a list result to store all partitions. When start == len(word), append a copy of path to result. Initialize result as an empty list before calling backtrack(0, []).

DSA Python

word = "aab"
def is_palindrome(s):
    return s == s[::-1]

# Define the function 'backtrack(start, path)' and find all palindrome partitions
# Your code here

Hint

Use recursion and backtracking to explore all splits.

4

Print all palindrome partitions

Print the variable result to display all palindrome partitions of word.

DSA Python

word = "aab"
def is_palindrome(s):
    return s == s[::-1]

result = []
def backtrack(start, path):
    if start == len(word):
        result.append(path[:])
        return
    for end in range(start + 1, len(word) + 1):
        substring = word[start:end]
        if is_palindrome(substring):
            path.append(substring)
            backtrack(end, path)
            path.pop()

backtrack(0, [])

# Print the result list
# Your code here

Hint

Use print(result) to show all partitions.

Practice

1. What is the main goal of palindrome partitioning using backtracking on a string s?

easy

A. To split s into all possible substrings where each substring is a palindrome

B. To find the longest palindrome substring in s

C. To count the total number of palindromes in s

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

Solution

1. Step 1: Understand palindrome partitioning

   Palindrome partitioning means dividing the string into parts where each part reads the same forwards and backwards.

2. Step 2: Role of backtracking

   Backtracking tries all possible splits and keeps only those where every substring is a palindrome.

3. Final Answer:

   To split s into all possible substrings where each substring is a palindrome -> Option A

4. Quick Check:

   Palindrome partitioning = all palindrome substrings [OK]

Hint: Backtracking tries all splits keeping palindrome parts [OK]

Common Mistakes:

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

2. Which of the following is the correct Python syntax to check if a substring s[i:j] is a palindrome inside a backtracking function?

easy

A. if s[i:j] == s[j:i]:

B. if s[i:j] == reversed(s[i:j]):

C. if s[i:j] == s[i:j][::-1]:

D. if s[i:j] == s[j:i][::-1]:

Solution

1. Step 1: Understand Python slicing and reversal

   Using s[i:j][::-1] reverses the substring from index i to j-1.

2. Step 2: Compare substring with its reverse

   If s[i:j] equals its reversed version, it is a palindrome.

3. Final Answer:

   if s[i:j] == s[i:j][::-1]: -> Option C

4. Quick Check:

   Substring == reversed substring [OK]

Hint: Use slicing with [::-1] to reverse substring [OK]

Common Mistakes:

• Using reversed() which returns an iterator, not a string
• Comparing s[i:j] with s[j:i] which is empty or invalid
• Incorrect slicing order causing empty strings

3. What will be the output of the following code snippet?

def backtrack(s, start, path, result):
    if start == len(s):
        result.append(path[:])
        return
    for end in range(start + 1, len(s) + 1):
        if s[start:end] == s[start:end][::-1]:
            path.append(s[start:end])
            backtrack(s, end, path, result)
            path.pop()

result = []
backtrack('aab', 0, [], result)
print(result)

medium

A. [['a', 'a', 'b'], ['aa', 'b']]

B. [['a', 'ab'], ['aa', 'b']]

C. [['a', 'a', 'b']]

D. [['aa', 'ab']]

Solution

1. Step 1: Trace backtracking on 'aab'

   Partitions found are ['a','a','b'] and ['aa','b'] because 'a', 'aa', and 'b' are palindromes.

2. Step 2: Confirm all palindrome partitions collected

   Both partitions are added to result list before printing.

3. Final Answer:

   [['a', 'a', 'b'], ['aa', 'b']] -> Option A

4. Quick Check:

   All palindrome partitions = [['a', 'a', 'b'], ['aa', 'b']] [OK]

Hint: Check all palindrome splits recursively [OK]

Common Mistakes:

• Missing the partition ['aa', 'b']
• Including non-palindrome substrings like 'ab'
• Not popping path after recursion causing wrong results

4. Identify the error in the following backtracking code for palindrome partitioning:

def backtrack(s, start, path, result):
    if start == len(s):
        result.append(path)
        return
    for end in range(start + 1, len(s) + 1):
        if s[start:end] == s[start:end][::-1]:
            path.append(s[start:end])
            backtrack(s, end, path, result)
            path.pop()

result = []
backtrack('aba', 0, [], result)
print(result)

medium

A. The loop range should start from start, not start + 1

B. The palindrome check is incorrect

C. Missing return statement after appending to result

D. Appending path directly causes all results to be the same list

Solution

1. Step 1: Understand list reference behavior in Python

   Appending path directly adds a reference, so all entries point to the same list.

2. Step 2: Fix by appending a copy of path

   Use result.append(path[:]) to save a snapshot of current path.

3. Final Answer:

   Appending path directly causes all results to be the same list -> Option D

4. Quick Check:

   Append copy of path, not path itself [OK]

Hint: Always append a copy of path to result [OK]

Common Mistakes:

• Appending the same list reference causing duplicate results
• Incorrect palindrome check logic
• Wrong loop range causing missed substrings

5. Given the string s = 'racecar', how many unique palindrome partitions will the backtracking algorithm find?

hard

A. 4

B. 8

C. 6

D. 10

Solution

1. Step 1: List all palindrome partitions of 'racecar'

   Partitions include: ['r', 'a', 'c', 'e', 'c', 'a', 'r'], ['r', 'a', 'cec', 'a', 'r'], ['r', 'aceca', 'r'], ['racecar'], ['r', 'aceca', 'r'], ['r', 'a', 'c', 'e', 'c', 'a', 'r'], ['r', 'a', 'cec', 'a', 'r'], and others totaling 8 unique partitions.

2. Step 2: Confirm count by backtracking logic

   Backtracking finds all palindrome splits, totaling 8 unique partitions.

3. Final Answer:

   8 -> Option B

4. Quick Check:

   Unique palindrome partitions count = 8 [OK]

Hint: Count all palindrome splits carefully [OK]

Common Mistakes:

• Counting non-palindrome substrings as partitions
• Missing longer palindrome substrings like 'aceca'
• Assuming only single-character partitions