Palindrome Partitioning Using Backtracking in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Palindrome Partitioning for "aab"

What is the output of the following C code that prints all palindrome partitions of the string "aab"?

DSA C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

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

void backtrack(char *s, int start, int len, char result[][100], int resultSize) {
    if (start == len) {
        for (int i = 0; i < resultSize; i++) {
            printf("%s", result[i]);
            if (i != resultSize - 1) printf("|");
        }
        printf("\n");
        return;
    }

    for (int end = start; end < len; end++) {
        if (isPalindrome(s, start, end)) {
            int substrLen = end - start + 1;
            strncpy(result[resultSize], s + start, substrLen);
            result[resultSize][substrLen] = '\0';
            backtrack(s, end + 1, len, result, resultSize + 1);
        }
    }
}

int main() {
    char s[] = "aab";
    int len = strlen(s);
    char result[100][100];
    backtrack(s, 0, len, result, 0);
    return 0;
}

A

a|ab
aa|b

B

a|a|b
aab

C

a|a|b
aa|b

D

aa|b
a|b

💡 Hint

Check all substrings that are palindromes and print partitions separated by '|'.

The code finds all palindrome partitions of "aab". The partitions are: "a|a|b" and "aa|b".

🧠 Conceptual

intermediate

Number of Palindrome Partitions for "abc"

How many palindrome partitions does the string "abc" have?

A 0

B 1

C 2

D 3

💡 Hint

Each single character is a palindrome itself.

The only palindrome partition is "a|b|c", since no consecutive characters form a palindrome. Thus, there is 1 palindrome partition.

❓ Predict Output

advanced

Output of Palindrome Partitioning for "racecar"

What is the output of the following C code that prints all palindrome partitions of the string "racecar"?

DSA C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

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

void backtrack(char *s, int start, int len, char result[][100], int resultSize) {
    if (start == len) {
        for (int i = 0; i < resultSize; i++) {
            printf("%s", result[i]);
            if (i != resultSize - 1) printf("|");
        }
        printf("\n");
        return;
    }

    for (int end = start; end < len; end++) {
        if (isPalindrome(s, start, end)) {
            int substrLen = end - start + 1;
            strncpy(result[resultSize], s + start, substrLen);
            result[resultSize][substrLen] = '\0';
            backtrack(s, end + 1, len, result, resultSize + 1);
        }
    }
}

int main() {
    char s[] = "racecar";
    int len = strlen(s);
    char result[100][100];
    backtrack(s, 0, len, result, 0);
    return 0;
}

A

r|a|c|e|c|a|r
r|a|cec|a|r
r|aceca|r
racecar

B

r|a|c|e|c|a|r
ra|cec|ar
racecar

C

r|a|c|e|c|a|r
r|aceca|r
racecar

D

racecar
r|a|c|e|c|a|r
r|a|cec|a|r

💡 Hint

Look for all palindrome substrings including single letters and longer palindromes like "cec", "aceca", and "racecar" itself.

The code prints all palindrome partitions of "racecar". The partitions include all single letters, and longer palindromes like "cec", "aceca", and the full "racecar".

🔧 Debug

advanced

Identify the error in palindrome check function

What error will the following palindrome check function cause when called with start > end?

DSA C

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

// Called as isPalindrome(s, 3, 2);

A Causes infinite loop

B Returns false without error

C Causes out-of-bounds access

D Returns true without error

💡 Hint

Check the while loop condition and what happens if start is greater than end initially.

If start > end initially, the while loop condition is false, so the function returns true immediately. This is correct behavior for empty or invalid substring.

🚀 Application

expert

Minimum Cuts for Palindrome Partitioning

Given the string "ababbbabbababa", what is the minimum number of cuts needed to partition it into palindromes?

A 3

B 5

C 4

D 6

💡 Hint

Use dynamic programming or backtracking to find minimum cuts for palindrome partitions.

The minimum cuts needed for "ababbbabbababa" is 3. One optimal partition is: "a|babbbab|b|ababa".

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