Palindrome Detection in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a string is a palindrome by comparing characters from start and end.

DSA C++

bool isPalindrome(const std::string& s) {
    int left = 0;
    int right = s.size() - 1;
    while (left < right) {
        if (s[left] != s[1]) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

Drag options to blanks, or click blank then click option'

A[right]

B(right)

C.at(right)

D[left]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets for indexing.

Comparing the character at left with itself instead of right.

The code compares characters at positions left and right using array indexing []. Using s[right] accesses the character at the right index.

2fill in blank

medium

Complete the code to convert a string to lowercase before checking palindrome.

DSA C++

std::string toLowerCase(const std::string& s) {
    std::string result = s;
    for (int i = 0; i < [1]; i++) {
        if (result[i] >= 'A' && result[i] <= 'Z') {
            result[i] = result[i] + 32;
        }
    }
    return result;
}

Drag options to blanks, or click blank then click option'

Aresult.size()

Bs.length()

Cresult.length()

Ds.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using a fixed number instead of the string length.

Using the size of the result string which is the same but less clear.

The loop should run from 0 to the size of the string. Using s.size() returns the length of the input string.

3fill in blank

hard

Fix the error in the palindrome check function by completing the blank.

DSA C++

bool isPalindrome(const std::string& s) {
    int left = 0;
    int right = s.size() - 1;
    while (left < right) {
        if (tolower(s[left]) != tolower(s[1])) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

Drag options to blanks, or click blank then click option'

A[right]

B(right)

C.at(right)

D[left]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets for indexing.

Comparing the character at left with itself.

The function compares characters at left and right indices after converting them to lowercase. Using s[right] accesses the correct character.

4fill in blank

hard

Fill both blanks to create a palindrome check ignoring non-alphanumeric characters.

DSA C++

bool isPalindrome(const std::string& s) {
    int left = 0;
    int right = s.size() - 1;
    while (left < right) {
        while (left < right && !isalnum(s[1])) left++;
        while (left < right && !isalnum(s[2])) right--;
        if (tolower(s[left]) != tolower(s[right])) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

Drag options to blanks, or click blank then click option'

A[left]

B[right]

C(left)

D(right)

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets for indexing.

Mixing up left and right indices.

To skip non-alphanumeric characters, we check s[left] and s[right] using square brackets inside the isalnum function.

5fill in blank

hard

Fill all three blanks to create a palindrome check that ignores case and non-alphanumeric characters.

DSA C++

bool isPalindrome(const std::string& s) {
    int left = 0;
    int right = s.size() - 1;
    while (left < right) {
        while (left < right && !isalnum(s[1])) [2]++;
        while (left < right && !isalnum(s[3])) right--;
        if (tolower(s[left]) != tolower(s[right])) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

Drag options to blanks, or click blank then click option'

A[left]

Bleft

C[right]

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets for indexing.

Incrementing or decrementing the wrong variable.

The code skips non-alphanumeric characters by checking s[left] and s[right] with square brackets, and increments or decrements the left and right indices accordingly.

Practice

1. What does a palindrome string mean in simple terms?

easy

A. A string that has all unique characters

B. A string that reads the same forwards and backwards

C. A string that contains only vowels

D. A string that ends with a special character

Solution

1. Step 1: Understand palindrome definition

   A palindrome is a word or phrase that reads the same from left to right and right to left.

2. Step 2: Compare options with definition

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

3. Final Answer:

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

4. Quick Check:

   Palindrome = same forwards and backwards [OK]

Hint: Palindrome means same forwards and backwards [OK]

Common Mistakes:

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

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

easy

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

B. for (int i = 0; i < s.size(); i++) if (s[i] == s[i + 1]) return true;

C. for (int i = 0; i < s.size(); i++) if (s[i] != s[i + 1]) return false;

D. for (int i = 1; i <= s.size(); i++) if (s[i] != s[s.size() - i]) return false;

Solution

1. Step 1: Analyze loop and index usage

   for (int i = 0; i < s.size() / 2; i++) if (s[i] != s[s.size() - i - 1]) return false; correctly loops till half the string and compares characters from start and end.

2. Step 2: Check correctness of index calculations

   for (int i = 0; i < s.size() / 2; i++) if (s[i] != s[s.size() - i - 1]) return false; uses s.size() - i - 1 which is correct for zero-based indexing.

3. Final Answer:

   for (int i = 0; i < s.size() / 2; i++) if (s[i] != s[s.size() - i - 1]) return false; -> Option A

4. Quick Check:

   Correct index and loop range = for (int i = 0; i < s.size() / 2; i++) if (s[i] != s[s.size() - i - 1]) return false; [OK]

Hint: Check characters from start and end up to middle [OK]

Common Mistakes:

• Using wrong loop range (full length instead of half)
• Incorrect index calculation causing out-of-range errors
• Comparing adjacent characters instead of mirrored ones

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

std::string s = "racecar";
bool isPalindrome = true;
for (int i = 0; i < s.size() / 2; i++) {
  if (s[i] != s[s.size() - i - 1]) {
    isPalindrome = false;
    break;
  }
}
std::cout << (isPalindrome ? "Yes" : "No") << std::endl;

medium

A. Yes

B. No

C. Compilation error

D. Runtime error

Solution

1. Step 1: Understand the string and loop

   The string "racecar" is a palindrome because it reads the same forwards and backwards.

2. Step 2: Trace the loop comparisons

   The loop compares characters from start and end; all pairs match, so isPalindrome remains true.

3. Final Answer:

   Yes -> Option A

4. Quick Check:

   "racecar" is palindrome = Yes [OK]

Hint: Check mirrored characters; all match means palindrome [OK]

Common Mistakes:

• Assuming "racecar" is not palindrome
• Confusing output strings
• Ignoring break statement in loop

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

bool checkPalindrome(std::string s) {
  for (int i = 0; i <= s.size() / 2; i++) {
    if (s[i] != s[s.size() - i]) {
      return false;
    }
  }
  return true;
}

medium

A. Comparison should be s[i] == s[s.size() - i]

B. No error, code works fine

C. Function should return void instead of bool

D. End index should be s[s.size() - i - 1] instead of s[s.size() - i] to avoid out-of-range

Solution

1. Step 1: Check loop range and indexing

   The end index s[s.size() - i] for i=0 accesses s[s.size()], causing out-of-range error.

2. Step 2: Correct index calculation

   Change to s[s.size() - i - 1] for proper zero-based end index mirroring.

3. Final Answer:

   End index should be s[s.size() - i - 1] instead of s[s.size() - i] to avoid out-of-range -> Option D

4. Quick Check:

   End index = size - i - 1 [OK]

Hint: End index: s.size() - i - 1 [OK]

Common Mistakes:

• Wrong end index s[size - i] causes out-of-bound access
• Confusing equality and inequality in comparisons
• Ignoring return type correctness

5. You want to check if a sentence is a palindrome ignoring spaces, punctuation, and case. Which approach is best?

hard

A. Only convert to lowercase and check palindrome

B. Check palindrome directly on original string without changes

C. Remove non-alphanumeric characters, convert to lowercase, then check palindrome

D. Reverse the string and compare without any preprocessing

Solution

1. Step 1: Understand palindrome ignoring spaces and punctuation

   To ignore spaces and punctuation, remove all non-alphanumeric characters and convert to lowercase for uniform comparison.

2. Step 2: Apply palindrome check on cleaned string

   After cleaning, check if the string reads the same forwards and backwards.

3. Final Answer:

   Remove non-alphanumeric characters, convert to lowercase, then check palindrome -> Option C

4. Quick Check:

   Clean string then check palindrome [OK]

Hint: Clean string first, then check palindrome [OK]

Common Mistakes:

• Checking palindrome without cleaning string
• Ignoring case differences
• Assuming spaces and punctuation don't affect palindrome