Palindrome Partitioning Using Backtracking in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of palindrome partitioning?

To split a string into parts where each part is a palindrome.

beginner

What does backtracking help with in palindrome partitioning?

It helps explore all possible ways to split the string and find all palindrome partitions.

beginner

How do you check if a substring is a palindrome?

Compare characters from start and end moving towards the center; if all match, it is a palindrome.

intermediate

Why do we remove the last added substring during backtracking?

To try other possible partitions by undoing the last choice and exploring new paths.

intermediate

What is the base case in palindrome partitioning backtracking?

When the start index reaches the end of the string, meaning a complete partition is found.

What is a palindrome?

A A string with all unique characters

B A string that reads the same forwards and backwards

C A string with only vowels

D A string with no repeating characters

A palindrome reads the same forwards and backwards.

In backtracking, what do you do after exploring one partition path?

A Stop and return the result

B Add more characters to the last partition

C Remove the last partition and try another

D Reverse the entire string

Backtracking removes the last choice to explore other possibilities.

Which of these substrings is a palindrome?

A abc

B acb

C ab

D aba

"aba" reads the same forwards and backwards.

What triggers the base case in palindrome partitioning backtracking?

A When the start index equals string length

B When the string has only one character

C When a non-palindrome substring is found

D When the string is empty

The base case is reached when the start index reaches the end of the string.

Why do we check palindrome before adding a substring to the partition?

A To ensure all parts are palindromes

B To reduce string length

C To sort the string

D To count vowels

Only palindrome substrings are valid parts in the partition.

Explain how backtracking is used to find all palindrome partitions of a string.

Think about choosing substrings and undoing choices to explore all options.

Describe the steps to check if a substring is a palindrome in the context of partitioning.

Imagine reading the substring from both ends at the same time.

Practice

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

easy

A. To find the longest palindrome substring in s

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

C. To count the number of palindromes in s

D. To reverse the string s using recursion

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

3. Final Answer:

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

4. Quick Check:

   Palindrome partitioning = all 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 string

2. Which of the following is the correct base case condition in a backtracking function for palindrome partitioning of string s starting at index start?

easy

A. if (start == s.length()) add current partition to result and return;

B. if (start > s.length()) return false;

C. if (start == s.length()) return true;

D. if (start < 0) return;

Solution

1. Step 1: Identify when to stop recursion

   When start reaches the length of the string, it means the entire string is partitioned.

2. Step 2: What to do at base case

   At this point, add the current list of palindrome substrings to the result list and return to explore other partitions.

3. Final Answer:

   if (start == s.length()) add current partition to result and return; -> Option A

4. Quick Check:

   Base case adds partition when start reaches string end [OK]

Hint: Add partition when start equals string length [OK]

Common Mistakes:

• Returning true/false instead of adding partition
• Checking start > length instead of == length
• Using invalid index checks like start < 0

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

medium

A. [["aab"]]

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

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

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

Solution

1. Step 1: Check all palindrome substrings

   "a" is palindrome, "aa" is palindrome, "b" is palindrome, but "ab" and "aab" are 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:

   Valid palindrome partitions = [["a", "a", "b"], ["aa", "b"]] [OK]

Hint: Check substrings palindrome before partitioning [OK]

Common Mistakes:

• Including non-palindrome substrings like "ab"
• Missing the partition ["aa", "b"]
• Returning the whole string as one partition

4. Given the following code snippet for palindrome partitioning, what is the error?

void backtrack(char* s, int start, char** current, int currentSize, char*** result, int* returnSize) {
  if (start == strlen(s)) {
    result[*returnSize] = current;
    (*returnSize)++;
    return;
  }
  for (int end = start; end < strlen(s); end++) {
    if (isPalindrome(s, start, end)) {
      current[currentSize] = substring(s, start, end);
      backtrack(s, end + 1, current, currentSize + 1, result, returnSize);
    }
  }
}

medium

A. Assigning current directly to result causes shared reference issues

B. The base case condition is incorrect

C. The loop should run from end = 0 to end < strlen(s)

D. Missing return statement after recursive call

Solution

1. Step 1: Analyze base case assignment

   Assigning current pointer directly to result[*returnSize] stores the same reference for all partitions.

2. Step 2: Understand backtracking data handling

   We must copy the current partition array to avoid overwriting when backtracking continues.

3. Final Answer:

   Assigning current directly to result causes shared reference issues -> Option A

4. Quick Check:

   Copy partitions to result to avoid shared references [OK]

Hint: Copy current partition before adding to result [OK]

Common Mistakes:

• Not copying current partition array
• Incorrect loop range for end index
• Ignoring base case correctness

5. How can you optimize palindrome checking in backtracking palindrome partitioning for string s of length n?

hard

A. Skip palindrome checking and add all substrings to partitions

B. Check palindrome by reversing substring every time during recursion

C. Precompute a 2D boolean table dp where dp[i][j] is true if substring s[i..j] is palindrome

D. Use a hash map to store all substrings and their palindrome status

Solution

1. Step 1: Understand palindrome checking cost

   Checking palindrome by reversing substring each time costs O(n) per check, leading to O(n^3) overall.

2. Step 2: Use dynamic programming optimization

   Precompute a 2D table dp where dp[i][j] stores palindrome status for substring s[i..j]. This reduces palindrome checks to O(1).

3. Step 3: Benefit in backtracking

   Using dp speeds up backtracking by avoiding repeated palindrome checks.

4. Final Answer:

   Precompute a 2D boolean table dp where dp[i][j] is true if substring s[i..j] is palindrome -> Option C

5. Quick Check:

   DP table for palindrome checking optimizes backtracking [OK]

Hint: Use DP table to check palindrome in O(1) time [OK]

Common Mistakes:

• Reversing substring every time causing slow performance
• Using hash map adds unnecessary overhead
• Skipping palindrome checks leads to wrong partitions