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.