Palindrome Detection in DSA C - Time & Space Complexity

Time Complexity: Palindrome Detection

O(n)

Understanding Time Complexity

We want to know how the time needed to check if a word or phrase is a palindrome changes as the input gets longer.

The question is: how does the number of steps grow when the input size grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <stdio.h>
#include <string.h>

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

This code checks if a string reads the same forwards and backwards by comparing characters from both ends moving towards the center.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing characters at positions from the start and end of the string.
• How many times: Up to half the length of the string (n/2 times).

How Execution Grows With Input

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

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

Pattern observation: The number of steps grows linearly with input size, but only half as many comparisons are needed.

Final Time Complexity

Time Complexity: O(n)

This means the time to check a palindrome grows in direct proportion to the length of the input string.

Common Mistake

[X] Wrong: "The palindrome check takes constant time because it just compares characters."

[OK] Correct: The number of comparisons depends on the string length, so longer strings take more time.

Interview Connect

Understanding how palindrome detection scales helps you explain your code clearly and shows you can analyze efficiency, a key skill in interviews.

Self-Check

"What if we checked only every other character instead of all? How would the time complexity change?"

Practice

1. What is a palindrome in the context of strings?

easy

A. A string that contains only vowels

B. A string that ends with a special character

C. A string with all unique characters

D. A string that reads the same forwards and backwards

Solution

1. Step 1: Understand palindrome definition

   A palindrome is a word or phrase that reads the same forwards and backwards, ignoring spaces and punctuation.

2. Step 2: Compare options with definition

   Only A string that reads the same forwards and backwards matches this definition exactly.

3. Final Answer:

   A string that reads the same forwards and backwards -> Option D

4. Quick Check:

   Palindrome = same forwards and backwards [OK]

Hint: Check if string equals its reverse [OK]

Common Mistakes:

• Confusing palindrome with unique characters
• Thinking palindrome means only vowels
• Assuming palindrome ends with special characters

2. Which of the following C code snippets correctly checks if a string s is a palindrome?

easy

A. for (int i = 0; i < strlen(s)/2; i++) { if (s[i] != s[strlen(s)-1-i]) return 0; } return 1;

B. for (int i = 0; i < strlen(s); i++) { if (s[i] == s[i+1]) return 1; } return 0;

C. for (int i = 0; i < strlen(s); i++) { if (s[i] != s[i+1]) return 0; } return 1;

D. for (int i = 0; i < strlen(s)/2; i++) { if (s[i] == s[strlen(s)-1-i]) return 0; } return 1;

Solution

1. Step 1: Analyze palindrome check logic

   We compare characters from start and end moving towards center. If any pair differs, string is not palindrome.

2. Step 2: Match code with logic

   for (int i = 0; i < strlen(s)/2; i++) { if (s[i] != s[strlen(s)-1-i]) return 0; } return 1; correctly compares s[i] and s[length-1-i] for half the string length and returns 0 if mismatch found, else 1.

3. Final Answer:

   for (int i = 0; i < strlen(s)/2; i++) { if (s[i] != s[strlen(s)-1-i]) return 0; } return 1; -> Option A

4. Quick Check:

   Compare front and back chars for half length [OK]

Hint: Check characters from ends moving inward [OK]

Common Mistakes:

• Comparing adjacent characters instead of opposite ends
• Returning 1 too early inside loop
• Incorrect loop range causing out-of-bounds

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

char s[] = "radar";
int i, flag = 1;
int n = strlen(s);
for (i = 0; i < n/2; i++) {
  if (s[i] != s[n - i - 1]) {
    flag = 0;
    break;
  }
}
printf("%d", flag);

medium

A. 0

B. 1

C. Compilation error

D. Undefined behavior

Solution

1. Step 1: Understand the code logic

   The code checks if string "radar" is palindrome by comparing characters from start and end.

2. Step 2: Trace the loop comparisons

   Characters match at all pairs: r-r, a-a, d-d (middle ignored). So flag remains 1.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   "radar" is palindrome, flag = 1 [OK]

Hint: Check if string equals reverse, print 1 if yes [OK]

Common Mistakes:

• Assuming flag is 0 for palindrome
• Confusing loop range causing missed checks
• Misreading string length calculation

4. Identify the error in this palindrome check code snippet:

int isPalindrome(char *s) {
  int n = strlen(s);
  for (int i = 0; i <= n/2; i++) {
    if (s[i] != s[n - i]) {
      return 0;
    }
  }
  return 1;
}

medium

A. Loop condition should be i < n/2, not i <= n/2

B. Function should return void instead of int

C. Index n - i should be n - i - 1 to avoid out-of-bounds

D. No error, code works correctly

Solution

1. Step 1: Check array indexing in condition

   String indices go from 0 to n-1, so s[n - i] accesses out-of-bounds when i=0.

2. Step 2: Correct index to n - i - 1

   Changing to s[n - i - 1] fixes indexing and prevents runtime error.

3. Final Answer:

   Index n - i should be n - i - 1 to avoid out-of-bounds -> Option C

4. Quick Check:

   Array index must be within 0 to n-1 [OK]

Hint: Remember last index is length - 1 [OK]

Common Mistakes:

• Using wrong loop boundary causing extra iteration
• Ignoring off-by-one in array indexing
• Assuming function return type is incorrect

5. Given a string with mixed case and spaces, which approach correctly checks if it is a palindrome ignoring case and spaces?

char s[] = "A man a plan a canal Panama";

hard

A. Remove spaces and convert to lowercase, then check palindrome normally

B. Check palindrome directly without modifications

C. Only convert to lowercase, ignore spaces

D. Only remove spaces, ignore case

Solution

1. Step 1: Normalize string by removing spaces and converting to lowercase

   Spaces and case differences affect palindrome check, so remove spaces and convert all letters to lowercase.

2. Step 2: Perform palindrome check on normalized string

   After normalization, check if string reads same forwards and backwards.

3. Final Answer:

   Remove spaces and convert to lowercase, then check palindrome normally -> Option A

4. Quick Check:

   Normalize string before palindrome check [OK]

Hint: Clean string first: lowercase + no spaces [OK]

Common Mistakes:

• Ignoring spaces causing false negatives
• Ignoring case differences causing false negatives
• Checking palindrome on raw string directly