Palindrome Detection in DSA C - Build from Scratch

Palindrome Detection

📖 Scenario: Imagine you are building a simple program that checks if a word is a palindrome. A palindrome is a word that reads the same forwards and backwards, like "madam" or "racecar".

🎯 Goal: You will create a program that takes a word, checks if it is a palindrome, and prints the result.

📋 What You'll Learn

Create a character array with a specific word

Create an integer variable for the word length

Write a loop to check if the word is a palindrome

Print whether the word is a palindrome or not

💡 Why This Matters

🌍 Real World

Palindrome detection is useful in text processing, data validation, and coding challenges.

💼 Career

Understanding string manipulation and loops is fundamental for programming jobs and technical interviews.

1

Create the word to check

Create a character array called word and initialize it with the string "level".

DSA C

# Create the character array called word with "level"
// Your code here

Hint

Use char word[] = "level"; to create the word.

2

Calculate the length of the word

Create an integer variable called length and set it to the length of word using sizeof(word) - 1.

DSA C

#include 

char word[] = "level";

// Calculate the length of word
// Your code here

Hint

Use int length = sizeof(word) - 1; to get the length without the null character.

3

Check if the word is a palindrome

Create an integer variable is_palindrome and set it to 1. Then write a for loop with int i = 0; i < length / 2; i++ to compare word[i] and word[length - 1 - i]. If they are not equal, set is_palindrome to 0 and break the loop.

DSA C

#include 

char word[] = "level";
int length = sizeof(word) - 1;

// Check if word is palindrome
// Your code here

Hint

Use a loop to compare characters from start and end moving towards the center.

4

Print the result

Use printf to print "Palindrome" if is_palindrome is 1, otherwise print "Not Palindrome".

DSA C

#include 

char word[] = "level";
int length = sizeof(word) - 1;

int is_palindrome = 1;
for (int i = 0; i < length / 2; i++) {
    if (word[i] != word[length - 1 - i]) {
        is_palindrome = 0;
        break;
    }
}

// Print "Palindrome" if is_palindrome is 1 else print "Not Palindrome"
// Your code here

Hint

Use if (is_palindrome) to decide what to print.

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