Valid Palindrome Two Pointer in DSA C - Build from Scratch

Valid Palindrome Two Pointer

📖 Scenario: Imagine you are building a simple text checker that tells if a word or phrase reads the same backward as forward, ignoring spaces, punctuation, and letter case. This is called a palindrome.We will use a two-pointer technique to check this efficiently.

🎯 Goal: Build a program that uses two pointers to check if a given string is a palindrome, ignoring non-alphanumeric characters and case differences.

📋 What You'll Learn

Create a string variable with the exact value "A man, a plan, a canal: Panama"

Create two integer variables called left and right to point to the start and end of the string

Implement a loop that moves left and right pointers towards the center, skipping non-alphanumeric characters

Compare characters at left and right ignoring case

Print "true" if the string is a palindrome, otherwise print "false"

💡 Why This Matters

🌍 Real World

Palindrome checks are used in text processing, data validation, and coding challenges.

💼 Career

Understanding two-pointer techniques and string manipulation is important for software development and technical interviews.

1

Create the input string

Create a character array called s and initialize it with the exact string "A man, a plan, a canal: Panama".

DSA C

# Create the string variable s with the given value
// Your code here

Hint

Use char s[] = "A man, a plan, a canal: Panama"; to create the string.

2

Initialize two pointers

Add two integer variables called left and right. Set left to 0 and right to the index of the last character in s (use strlen(s) - 1). Include #include <string.h> at the top.

DSA C

#include 
#include 

int main() {
    char s[] = "A man, a plan, a canal: Panama";
    // Initialize left and right pointers
    // Your code here
    return 0;
}

Hint

Use int left = 0; and int right = strlen(s) - 1; to set pointers.

3

Implement the two-pointer palindrome check

Write a while loop that runs while left < right. Inside the loop, skip non-alphanumeric characters by moving left forward and right backward. Use isalnum() from <ctype.h>. Compare characters at left and right ignoring case using tolower(). If they differ, break the loop. Otherwise, move left forward and right backward.

DSA C

#include 
#include 
#include 

int main() {
    char s[] = "A man, a plan, a canal: Panama";
    int left = 0;
    int right = strlen(s) - 1;
    // Implement two-pointer palindrome check
    // Your code here
    return 0;
}

Hint

Use nested while loops to skip non-alphanumeric characters and compare characters with tolower().

4

Print the result

After the loop, print "true" if the string is a palindrome (use the variable is_palindrome), otherwise print "false". Use printf().

DSA C

#include 
#include 
#include 

int main() {
    char s[] = "A man, a plan, a canal: Panama";
    int left = 0;
    int right = strlen(s) - 1;
    int is_palindrome = 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])) {
            is_palindrome = 0;
            break;
        }
        left++;
        right--;
    }
    // Print "true" if palindrome, else "false"
    // Your code here
    return 0;
}

Hint

Use printf("true\n"); or printf("false\n"); based on is_palindrome.

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