Valid Palindrome Two Pointer in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Two Pointer Palindrome Check

What is the output of the following code when checking if the string "A man, a plan, a canal: Panama" is a valid palindrome?

DSA C

bool isPalindrome(char * s) {
    int left = 0, right = strlen(s) - 1;
    while (left < right) {
        while (left < right && !isalnum(s[left])) left++;
        while (left < right && !isalnum(s[right])) right--;
        if (tolower(s[left]) != tolower(s[right])) return false;
        left++;
        right--;
    }
    return true;
}

int main() {
    char s[] = "A man, a plan, a canal: Panama";
    printf("%s\n", isPalindrome(s) ? "true" : "false");
    return 0;
}

A Runtime error

B true

C Compilation error

D false

💡 Hint

Consider only alphanumeric characters and ignore cases.

The code uses two pointers moving inward, skipping non-alphanumeric characters and comparing lowercase characters. The input string is a classic palindrome ignoring spaces and punctuation.

❓ Predict Output

intermediate

Output for Non-Palindrome String

What will the output be when the code checks if the string "race a car" is a valid palindrome?

DSA C

bool isPalindrome(char * s) {
    int left = 0, right = strlen(s) - 1;
    while (left < right) {
        while (left < right && !isalnum(s[left])) left++;
        while (left < right && !isalnum(s[right])) right--;
        if (tolower(s[left]) != tolower(s[right])) return false;
        left++;
        right--;
    }
    return true;
}

int main() {
    char s[] = "race a car";
    printf("%s\n", isPalindrome(s) ? "true" : "false");
    return 0;
}

A false

B true

C Compilation error

D Runtime error

💡 Hint

Check characters ignoring spaces and punctuation carefully.

The string "race a car" is not a palindrome when ignoring spaces and punctuation because 'e' and 'a' do not match at the corresponding positions.

🧠 Conceptual

advanced

Why Two Pointers Work for Palindrome Check

Why is the two-pointer technique efficient for checking if a string is a palindrome?

A It converts the string to uppercase before checking.

B It sorts the string first to check for palindrome properties.

C It uses extra memory to store reversed string for comparison.

D It compares characters from both ends simultaneously, reducing the number of comparisons.

💡 Hint

Think about how palindrome characters relate from start and end.

Two pointers start from the beginning and end and move inward, comparing characters directly without extra space or full string reversal.

🔧 Debug

advanced

Identify the Bug in Palindrome Code

What is the bug in this palindrome checking code snippet?

DSA C

bool isPalindrome(char * s) {
    int left = 0, right = strlen(s);
    while (left < right) {
        if (tolower(s[left]) != tolower(s[right])) return false;
        left++;
        right--;
    }
    return true;
}

A right should be initialized to strlen(s) - 1 to avoid out-of-bounds access.

B tolower should not be used on characters.

C left and right pointers should both start at 0.

D The loop condition should be while (left <= right).

💡 Hint

Check the index of the last character in the string.

strlen(s) returns the length, but valid indices go from 0 to length-1. Using length as index causes out-of-bounds access.

🚀 Application

expert

Count Palindromic Substrings Using Two Pointers

Given a string, which approach correctly counts all palindromic substrings using the two-pointer technique?

A Use a hash map to store all substrings and check each for palindrome.

B Reverse the string and compare substrings directly.

C Expand around each center (single and between characters) using two pointers to count palindromes.

D Sort the string and count repeated characters as palindromes.

💡 Hint

Think about how to find palindromes centered at each position.

Expanding around centers with two pointers allows counting all palindromic substrings efficiently without extra space.