Palindrome Detection in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a palindrome?

A palindrome is a word, number, phrase, or sequence that reads the same backward as forward, like 'madam' or '121'.

beginner

How can you check if a string is a palindrome using two pointers?

Use two pointers: one at the start and one at the end of the string. Move them towards the center, comparing characters. If all pairs match, it's a palindrome.

intermediate

Why is ignoring case important in palindrome detection?

Ignoring case ensures that uppercase and lowercase letters are treated the same, so 'Madam' is recognized as a palindrome just like 'madam'.

intermediate

What is the time complexity of checking a palindrome in a string of length n?

The time complexity is O(n) because each character is checked at most once when comparing pairs from start and end.

beginner

Show a simple C code snippet to check if a string is a palindrome.

#include #include #include int isPalindrome(char str[]) { int left = 0; int right = strlen(str) - 1; while (left < right) { if (tolower(str[left]) != tolower(str[right])) { return 0; // Not palindrome } left++; right--; } return 1; // Palindrome } int main() { char s[] = "Madam"; if (isPalindrome(s)) { printf("Palindrome\n"); } else { printf("Not Palindrome\n"); } return 0; }

What does a palindrome string read the same as?

A Only backward

B Only forward

C Backward and forward

D Random order

A palindrome reads the same backward and forward.

Which method is commonly used to check palindrome in a string?

A Two pointers from start and end

B Sorting the string

C Reversing the string and comparing

D Counting characters

Two pointers from start and end compare characters to check palindrome.

What should you do to ignore case when checking palindrome?

A Convert characters to lowercase or uppercase before comparing

B Ignore spaces only

C Compare ASCII values directly

D Use only uppercase letters

Converting characters to the same case ensures case-insensitive comparison.

What is the time complexity of palindrome detection in a string of length n?

A O(n^2)

B O(1)

C O(log n)

D O(n)

Checking pairs from start and end requires O(n) time.

If the characters at the two pointers do not match, what can you conclude?

A Check the next characters

B The string is not a palindrome

C The string is a palindrome

D Ignore the mismatch

A mismatch means the string cannot be a palindrome.

Explain how to check if a string is a palindrome using the two-pointer technique.

Think about comparing characters from both ends moving inward.

Describe why ignoring case is important in palindrome detection and how to implement it in C.

Consider how 'Madam' and 'madam' should be treated the same.

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