Valid Palindrome Two Pointer in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the two-pointer technique in checking a palindrome?

Use two pointers starting at the beginning and end of the string, move them towards the center, and compare characters to check if they match.

beginner

Why do we skip non-alphanumeric characters in the Valid Palindrome problem?

Because only letters and numbers matter for palindrome checking; spaces, punctuation, and symbols are ignored.

beginner

How do you handle case differences when checking for a palindrome?

Convert all characters to the same case (usually lowercase) before comparing to ensure case-insensitive matching.

beginner

What happens if the characters at the two pointers do not match during the palindrome check?

The string is not a palindrome, so the function should return false immediately.

intermediate

What is the time complexity of the two-pointer palindrome check?

O(n), where n is the length of the string, because each character is checked at most once.

In the two-pointer palindrome check, where do the pointers start?

A In the middle of the string

B At the start and end of the string

C Both at the end

D Both at the start

The two pointers start at the beginning and end of the string and move towards the center.

Which characters are ignored in the Valid Palindrome problem?

A Alphanumeric characters

B Letters only

C Spaces and punctuation

D Numbers only

Spaces, punctuation, and other non-alphanumeric characters are ignored.

How do you compare characters to ensure case-insensitive palindrome checking?

A Compare as is

B Ignore case differences

C Convert both to uppercase

D Convert both to lowercase

Converting both characters to lowercase ensures case-insensitive comparison.

What should the function return if a mismatch is found between characters?

A False

B True

C Continue checking

D Skip characters

If characters do not match, the string is not a palindrome, so return false.

What is the overall time complexity of the two-pointer palindrome check?

A O(n)

B O(n^2)

C O(log n)

D O(1)

Each character is checked once, so the time complexity is O(n).

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

Think about comparing characters from both ends moving towards the center.

Describe how to handle characters that are not letters or numbers when checking for a palindrome.

Focus on ignoring spaces and punctuation.

Practice

1. What is the main idea behind the two-pointer technique to check if a string is a palindrome?

easy

A. Sort the string and check if it matches the original

B. Use two pointers starting from both ends and move towards the center comparing characters

C. Reverse the string and compare with the original

D. Count the frequency of each character and check if all are even

Solution

1. Step 1: Understand two-pointer approach

   Two pointers start at the beginning and end of the string and move inward.

2. Step 2: Compare characters at pointers

   If characters differ, string is not palindrome; if all match, it is palindrome.

3. Final Answer:

   Use two pointers starting from both ends and move towards the center comparing characters -> Option B

4. Quick Check:

   Two-pointer technique = 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 string instead of comparing characters
• Only checking half the string without pointers
• Ignoring case or non-alphanumeric characters

2. Which of the following C++ code snippets correctly initializes two pointers for palindrome checking on string s?

easy

A. int left = 0, right = s.size() - 1;

B. int left = 1, right = s.size();

C. int left = s.size(), right = 0;

D. int left = s.size() - 1, right = 0;

Solution

1. Step 1: Identify valid pointer initialization

   Left pointer starts at 0 (first char), right pointer at last index (size - 1).

2. Step 2: Check each option

   int left = 0, right = s.size() - 1; correctly sets left=0 and right=s.size()-1; others have invalid indices.

3. Final Answer:

   int left = 0, right = s.size() - 1; -> Option A

4. Quick Check:

   Pointers start at 0 and size-1 = int left = 0, right = s.size() - 1; [OK]

Hint: Left = 0, Right = length - 1 for string indices [OK]

Common Mistakes:

• Starting right pointer at s.size() (out of range)
• Swapping left and right initialization
• Starting left pointer at 1 instead of 0

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

std::string s = "A man, a plan, a canal: Panama";
int left = 0, right = s.size() - 1;
while (left < right) {
  while (left < right && !std::isalnum(s[left])) left++;
  while (left < right && !std::isalnum(s[right])) right--;
  if (std::tolower(s[left]) != std::tolower(s[right])) {
    std::cout << "Not Palindrome" << std::endl;
    return 0;
  }
  left++;
  right--;
}
std::cout << "Palindrome" << std::endl;

medium

A. Compilation Error

B. Not Palindrome

C. Runtime Error

D. Palindrome

Solution

1. Step 1: Understand the code logic

   It skips non-alphanumeric chars and compares characters case-insensitively from both ends.

2. Step 2: Check if string is palindrome ignoring cases and symbols

   "A man, a plan, a canal: Panama" is a classic palindrome ignoring spaces and punctuation.

3. Final Answer:

   Palindrome -> Option D

4. Quick Check:

   Ignoring non-alnum and case, string is palindrome = Palindrome [OK]

Hint: Ignore non-letters and compare lowercase chars [OK]

Common Mistakes:

• Not skipping non-alphanumeric characters
• Comparing characters case-sensitively
• Confusing output with runtime or compile errors

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

std::string s = "racecar";
int left = 0, right = s.size();
while (left < right) {
  if (s[left] != s[right]) {
    std::cout << "Not Palindrome" << std::endl;
    break;
  }
  left++;
  right--;
}
std::cout << "Palindrome" << std::endl;

medium

A. Right pointer initialized to s.size() which is out of range

B. Left pointer should start at 1

C. Missing tolower() for case-insensitive check

D. Loop condition should be left <= right

Solution

1. Step 1: Check pointer initialization

   Right pointer is set to s.size(), which is one past the last valid index.

2. Step 2: Understand consequences

   Accessing s[right] causes out-of-bounds error or undefined behavior.

3. Final Answer:

   Right pointer initialized to s.size() which is out of range -> Option A

4. Quick Check:

   Right pointer must be s.size()-1, not s.size() [OK]

Hint: Right pointer = length - 1 to avoid out-of-range [OK]

Common Mistakes:

• Using s.size() as last index
• Ignoring out-of-bounds access errors
• Assuming loop condition fixes index error

5. Given a string s, how would you modify the two-pointer palindrome check to ignore spaces, punctuation, and be case-insensitive? Choose the best approach.

hard

A. Convert string to uppercase and compare characters including spaces

B. Reverse the string and compare directly without any changes

C. Use two pointers; skip non-alphanumeric chars; compare lowercase characters

D. Count characters frequency and check if all are even

Solution

1. Step 1: Handle non-alphanumeric characters

   Skip spaces and punctuation by moving pointers until alphanumeric chars found.

2. Step 2: Compare characters case-insensitively

   Convert characters to lowercase before comparing to ignore case differences.

3. Final Answer:

   Use two pointers; skip non-alphanumeric chars; compare lowercase characters -> Option C

4. Quick Check:

   Skip non-alnum + lowercase compare = Use two pointers; skip non-alphanumeric chars; compare lowercase characters [OK]

Hint: Skip non-letters and compare lowercase chars [OK]

Common Mistakes:

• Comparing characters without skipping punctuation
• Ignoring case differences
• Using frequency count which doesn't check order