Valid Palindrome Two Pointer in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the left pointer at the start of the string.

DSA C

int left = [1];

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

A0

B1

Cstrlen(s)

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 skips the first character.

Using strlen(s) causes out-of-bounds access.

The left pointer should start at index 0 to begin checking from the start of the string.

2fill in blank

medium

Complete the code to initialize the right pointer at the end of the string.

DSA C

int right = [1] - 1;

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

A0

Bstrlen(s)

C1

Dsizeof(s)

Attempts:

3 left

💡 Hint

Common Mistakes

Using sizeof(s) returns size of pointer, not string length.

Setting right to 0 starts at the beginning, not the end.

The right pointer should start at the last character index, which is length minus one.

3fill in blank

hard

Fix the error in the while loop condition to continue while left is less than right.

DSA C

while ([1] < right) {

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

Aright - 1

Bright

Cleft + 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right causes no iterations.

Using left + 1 < right is incorrect syntax.

The loop should run while the left pointer is less than the right pointer.

4fill in blank

hard

Fill both blanks to skip non-alphanumeric characters from the left and right pointers.

DSA C

while (left < right && !isalnum(s[[1]])) left++;
while (left < right && !isalnum(s[[2]])) right--;

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

Aleft

Bright

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using fixed indices like 0 or 1 ignores pointer positions.

Mixing left and right in wrong places causes logic errors.

We check characters at the current left and right indices to skip non-alphanumeric characters.

5fill in blank

hard

Fill all three blanks to compare characters ignoring case and move pointers accordingly.

DSA C

if (tolower(s[[1]]) != tolower(s[[2]])) {
    return false;
}
[3];
[4];

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

Aleft

Bright

Cleft++

Dright--

Attempts:

3 left

💡 Hint

Common Mistakes

Not using tolower() causes case-sensitive mismatches.

Forgetting to move pointers causes infinite loops.

We compare characters at left and right ignoring case. If they differ, return false. Otherwise, move left forward and right backward.

Practice

1. What is the main idea behind the two-pointer technique to check if a string is a palindrome?

easy

A. Count the frequency of each character and check if all are even

B. Sort the string and check if it matches the original

C. Reverse the string and compare with the original

D. Use two pointers starting from both ends and move towards the center comparing characters

Solution

1. Step 1: Understand two-pointer setup

   Two pointers start at the beginning and end of the string to compare characters.

2. Step 2: Move pointers inward while characters match

   If characters at both pointers are equal, move pointers closer until they meet or cross.

3. Final Answer:

   Use two pointers starting from both ends and move towards the center comparing characters -> Option D

4. Quick Check:

   Two-pointer palindrome check = Use two pointers starting from both ends and move towards the center comparing characters [OK]

Hint: Compare characters from ends moving inward [OK]

Common Mistakes:

• Trying to sort the string which changes order
• Only checking half the string without pointers
• Ignoring non-alphanumeric characters if required

2. Which of the following C code snippets correctly initializes two pointers for palindrome checking on a string s?

easy

A. int left = 0, right = strlen(s) - 1;

B. int left = 1, right = strlen(s);

C. int left = 0, right = strlen(s);

D. int left = 1, right = strlen(s) - 1;

Solution

1. Step 1: Understand string indexing in C

   Strings are zero-indexed, so first character is at 0 and last at length-1.

2. Step 2: Correct pointer initialization

   Left pointer starts at 0, right pointer at strlen(s)-1 to cover full string.

3. Final Answer:

   int left = 0, right = strlen(s) - 1; -> Option A

4. Quick Check:

   Indexing starts at 0, last index = length-1 [OK]

Hint: Start left at 0 and right at length minus one [OK]

Common Mistakes:

• Setting right pointer to strlen(s) which is out of bounds
• Starting left pointer at 1 skipping first character
• Using wrong indices causing runtime errors

3. What will be the output of the following C code snippet?

char s[] = "racecar";
int left = 0, right = strlen(s) - 1;
int isPalindrome = 1;
while (left < right) {
    if (s[left] != s[right]) {
        isPalindrome = 0;
        break;
    }
    left++;
    right--;
}
printf("%d", isPalindrome);

medium

A. 0

B. 1

C. 7

D. Compilation error

Solution

1. Step 1: Trace pointer comparisons on "racecar"

   Characters at left and right match: 'r'=='r', 'a'=='a', 'c'=='c', 'e'=='e'.

2. Step 2: Loop completes without mismatch

   No mismatch found, so isPalindrome remains 1 and prints 1.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   "racecar" is palindrome = 1 [OK]

Hint: Check matching pairs from ends; no mismatch means palindrome [OK]

Common Mistakes:

• Assuming output is string instead of integer
• Confusing index bounds causing wrong comparisons
• Forgetting to break loop on mismatch

4. Identify the bug in the following C code that checks if a string s is a palindrome using two pointers:

int left = 0, right = strlen(s);
while (left <= right) {
    if (s[left] != s[right]) {
        return 0;
    }
    left++;
    right--;
}
return 1;

medium

A. Both A and B are bugs

B. Loop condition should be left < right, not left <= right

C. Right pointer should be initialized to strlen(s) - 1

D. No bug, code is correct

Solution

1. Step 1: Check right pointer initialization

   Right pointer set to strlen(s) is out of bounds; should be strlen(s) - 1.

2. Step 2: Check loop condition

   Loop should run while left < right to avoid comparing same middle character twice.

3. Final Answer:

   Both A and B are bugs -> Option A

4. Quick Check:

   Right pointer and loop condition both incorrect = Both A and B are bugs [OK]

Hint: Right pointer = length-1 and loop: left < right [OK]

Common Mistakes:

• Using strlen(s) as right pointer causing out-of-bounds access
• Using <= in loop causing unnecessary comparison
• Ignoring off-by-one errors

5. Given a string s that may contain uppercase letters, lowercase letters, and non-alphanumeric characters, which approach correctly checks if s is a palindrome ignoring cases and non-alphanumeric characters using two pointers?

hard

A. Count frequency of alphanumeric characters and check if all counts are even

B. Convert entire string to lowercase, remove non-alphanumeric chars, then compare with reversed string

C. Use two pointers from ends, skip non-alphanumeric chars, compare lowercase characters

D. Use two pointers but compare characters directly without skipping or case conversion

Solution

1. Step 1: Use two pointers and skip non-alphanumeric characters

   Move left and right pointers inward, skipping characters that are not letters or digits.

2. Step 2: Compare characters in lowercase form

   Convert characters to lowercase before comparing to ignore case differences.

3. Final Answer:

   Use two pointers from ends, skip non-alphanumeric chars, compare lowercase characters -> Option C

4. Quick Check:

   Two-pointer with skip and lowercase = Use two pointers from ends, skip non-alphanumeric chars, compare lowercase characters [OK]

Hint: Skip non-alphanumeric and compare lowercase chars with two pointers [OK]

Common Mistakes:

• Comparing characters without skipping non-alphanumeric
• Not converting characters to lowercase before comparison
• Using frequency count which doesn't check order