Valid Palindrome Two Pointer in DSA C - Time & Space Complexity

Time Complexity: Valid Palindrome Two Pointer

O(n)

Understanding Time Complexity

We want to know how the time to check if a string is a palindrome grows as the string gets longer.

Specifically, how many steps does the two-pointer method take as input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code checks if a string reads the same forwards and backwards using two pointers moving towards the center.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing characters at left and right pointers.
• How many times: Up to n/2 times, where n is the string length.

How Execution Grows With Input

As the string gets longer, the number of comparisons grows roughly half as fast.

Input Size (n)	Approx. Operations
10	5 comparisons
100	50 comparisons
1000	500 comparisons

Pattern observation: The steps grow linearly with input size, about half the length.

Final Time Complexity

Time Complexity: O(n)

This means the time to check grows directly with the string length.

Common Mistake

[X] Wrong: "The two-pointer method checks every character twice, so it is O(2n)."

[OK] Correct: Each character is checked only once in a pair, so the total checks are about n/2, which simplifies to O(n).

Interview Connect

Understanding this linear time check helps you quickly validate strings in many problems, showing you can write efficient, clear code.

Self-Check

"What if we ignored non-alphanumeric characters and spaces? How would the time complexity change?"

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