Longest Palindromic Substring in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the maximum length of palindrome found.

DSA C

int maxLength = [1];

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

A1

B-1

C0

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing maxLength to 0 which ignores single character palindromes.

Using NULL which is not an integer.

The minimum length of a palindrome substring is 1 (a single character).

2fill in blank

medium

Complete the code to expand around the center and check palindrome boundaries.

DSA C

while (left >= 0 && right < n && s[left] == s[right]) {
    left--;
    right++;
}
int length = right - left - [1];

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

A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Not subtracting 1 causes length to be off by 1.

Subtracting 0 or 2 leads to incorrect palindrome length.

After the while loop, left and right are one step beyond the palindrome boundaries, so subtract 1 to get the correct length.

3fill in blank

hard

Fix the error in the code that updates the start index of the longest palindrome found.

DSA C

if (length > maxLength) {
    start = left + [1];
    maxLength = length;
}

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

A-1

B1

C0

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using left directly without adjustment.

Adding 0 or 2 which shifts start incorrectly.

Since left was decremented one extra time in the expansion, add 1 to get the correct start index.

4fill in blank

hard

Fill both blanks to correctly check for odd and even length palindromes around center i.

DSA C

int len1 = expandAroundCenter(s, [1], [2]);
int len2 = expandAroundCenter(s, i, i + 1);

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

Ai

Bi - 1

Ci + 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using different indices for left and right in odd palindrome check.

Using i-1 or i+1 which is for even length palindrome.

For odd length palindrome, expand around the same center i (left and right both i).

5fill in blank

hard

Fill all three blanks to update start and maxLength after finding the longer palindrome length.

DSA C

int maxLen = (len1 > len2) ? len1 : len2;
if (maxLen > maxLength) {
    start = i - (maxLen - [1]) / [2];
    maxLength = maxLen;
    end = i + maxLen / [3];
}

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

A1

B2

C0

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Incorrect division causing wrong start or end indices.

Using 0 or 3 which do not represent half.

To find the start index, subtract half of (maxLen - 1). For end index, add half of maxLen.