Palindrome Partitioning Using Backtracking in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a substring is a palindrome.

DSA C

int isPalindrome(char *s, int start, int end) {
    while (start < end) {
        if (s[start] != s[end]) {
            return [1];
        }
        start++;
        end--;
    }
    return 1;
}

Drag options to blanks, or click blank then click option'

A0

B1

C-1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 when characters don't match.

Not returning any value inside the if condition.

Return 0 when characters at start and end do not match, meaning substring is not palindrome.

2fill in blank

medium

Complete the code to add a substring to the current partition list.

DSA C

void addSubstring(char *s, int start, int end, char result[][100], int *resIndex) {
    int i, j = 0;
    for (i = start; i <= end; i++) {
        result[*resIndex][j++] = s[i];
    }
    result[*resIndex][j] = '\0';
    (*resIndex)[1];
}

Drag options to blanks, or click blank then click option'

A++

B--

C+=

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using -- which decreases the index.

Using = which overwrites the index.

Increment the result index after adding a substring to move to next position.

3fill in blank

hard

Fix the error in the recursive backtracking call to explore partitions.

DSA C

void backtrack(char *s, int start, char result[][100], int resIndex) {
    if (s[start] == '\0') {
        // print or store the partition
        return;
    }
    for (int end = start; s[end] != '\0'; end++) {
        if (isPalindrome(s, start, end)) {
            addSubstring(s, start, end, result, &resIndex);
            backtrack(s, [1], result, resIndex);
        }
    }
}

Drag options to blanks, or click blank then click option'

Aend

Bstart + 1

Cend + 1

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using start + 1 which may skip substrings.

Using end which causes infinite recursion.

After choosing substring s[start..end], next recursion starts from end+1 to avoid overlap.

4fill in blank

hard

Fill both blanks to correctly backtrack after recursive call.

DSA C

void backtrack(char *s, int start, char result[][100], int *resIndex) {
    if (s[start] == '\0') {
        // print or store the partition
        return;
    }
    for (int end = start; s[end] != '\0'; end++) {
        if (isPalindrome(s, start, end)) {
            addSubstring(s, start, end, result, resIndex);
            backtrack(s, end + 1, result, resIndex);
            [1];
            [2];
        }
    }
}

Drag options to blanks, or click blank then click option'

A*resIndex--

B(*resIndex)--

CresIndex--

D(*resIndex)++

Attempts:

3 left

💡 Hint

Common Mistakes

Using resIndex-- which changes pointer address.

Not decrementing resIndex causing wrong partitions.

After recursion, decrement resIndex to remove last substring, then increment to restore for next iteration.

5fill in blank

hard

Fill all three blanks to initialize and start palindrome partitioning.

DSA C

int main() {
    char s[] = "aab";
    char result[100][100];
    int [1] = 0;
    backtrack(s, [2], result, &[3]);
    return 0;
}

Drag options to blanks, or click blank then click option'

AresIndex

B0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Passing resIndex instead of &resIndex.

Starting backtrack from 1 instead of 0.

Initialize resIndex to 0, start backtrack from index 0, and pass address of resIndex.

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