Palindrome Partitioning Using Backtracking in DSA C - 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.

bool isPalindrome(char* s, int start, int end) {
    while (start < end) {
        if (s[start] != s[end]) return false;
        start++;
        end--;
    }
    return true;
}

void backtrack(char* s, int start, int len) {
    if (start == len) {
        // store current partition
        return;
    }
    for (int end = start; end < len; end++) {
        if (isPalindrome(s, start, end)) {
            backtrack(s, end + 1, len);
        }
    }
}

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 for-loop tries every possible substring start to end.
• How many times: For each start position, it checks all ends, and calls backtrack recursively for the rest.
• Palindrome check: Each substring check scans characters between start and end.
• Dominant operation: The recursive calls combined with palindrome checks cause exponential growth.

How Execution Grows With Input

As the string length grows, the number of palindrome partitions grows very fast.

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 work grows exponentially because each position can split in many ways, and palindrome checks add extra cost.

Final Time Complexity

Time Complexity: O(n * 2^n)

This means the time grows very fast, roughly doubling with each added character, multiplied by the cost to check palindromes.

Common Mistake

[X] Wrong: "The algorithm runs in polynomial time because it just loops over the string."

[OK] Correct: The recursive splitting creates many branches, causing exponential growth, not just simple loops.

Interview Connect

Understanding this complexity helps you explain why backtracking can be slow and when to expect exponential time, a key skill in problem solving.

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 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