Palindrome Partitioning Using Backtracking in DSA Python - Time & Space Complexity

Time Complexity: Palindrome Partitioning Using Backtracking

O(n * 2^n)

Understanding Time Complexity

We want to understand how the time needed to find all palindrome partitions grows as the input string gets longer.

Specifically, how does the backtracking process scale with the string size?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def is_palindrome(s, start, end):
    while start < end:
        if s[start] != s[end]:
            return False
        start += 1
        end -= 1
    return True

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

This code finds all ways to split the string into palindrome parts using backtracking.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The recursive backtracking explores all substring partitions, checking each substring for palindrome.
• How many times: The recursion can branch many times, exploring many substring splits; palindrome checks happen for each substring considered.

How Execution Grows With Input

As the string length grows, the number of possible partitions grows very fast, and each partition requires palindrome checks.

Input Size (n)	Approx. Operations
10	Thousands of checks and recursive calls
100	Millions to billions of operations
1000	Operations grow beyond practical limits

Pattern observation: The operations grow faster than any simple polynomial, roughly exponentially with input size.

Final Time Complexity

Time Complexity: O(n * 2^n)

This means the time grows very fast as the string gets longer, because we explore many partitions and check palindromes for each.

Common Mistake

[X] Wrong: "The algorithm runs in polynomial time because it just uses loops and recursion."

[OK] Correct: The number of partitions grows exponentially, so the recursion explores an exponential number of paths, not just polynomial.

Interview Connect

Understanding this complexity helps you explain why backtracking can be expensive and when to expect exponential time, which is valuable for problem-solving discussions.

Self-Check

"What if we used memoization to store palindrome checks? How would the time complexity change?"

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