Valid Palindrome Two Pointer in DSA Javascript - 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 are the same.

beginner

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

Because the problem only cares about letters and numbers, ignoring spaces, punctuation, and symbols to focus on meaningful characters.

beginner

In the two-pointer palindrome check, what happens if characters at pointers don't match?

We conclude the string is not a palindrome and stop checking further.

beginner

How do you handle case sensitivity in the Valid Palindrome Two Pointer approach?

Convert all characters to the same case (usually lowercase) before comparing to ignore case differences.

intermediate

What is the time complexity of the two-pointer approach for checking a palindrome?

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

What do the two pointers represent in the Valid Palindrome Two Pointer method?

A Positions of vowels only

B Start and end positions of the string

C Random positions in the string

D Positions of spaces

The two pointers start at the beginning and end of the string to compare characters moving inward.

Which characters are ignored when checking for a valid palindrome?

A All characters are checked

B Only uppercase letters

C Numbers only

D Non-alphanumeric characters like spaces and punctuation

Non-alphanumeric characters are skipped to focus on letters and numbers only.

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

A Return false immediately

B Move both pointers inward

C Skip the characters and continue

D Convert characters to uppercase

A mismatch means the string is not a palindrome, so return false immediately.

How do you handle uppercase and lowercase letters in this approach?

A Ignore case by converting all to lowercase

B Compare as is without conversion

C Convert all to uppercase only

D Skip uppercase letters

Converting all characters to lowercase ensures case-insensitive comparison.

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

A O(n^2)

B O(1)

C O(n)

D O(log n)

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

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

Think about how you would check a word from both ends moving towards the center.

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

Imagine ignoring spaces and punctuation when reading a sentence backwards.

Practice

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

easy

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

B. Reverse the string and compare it with the original string

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

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

Solution

1. Step 1: Understand the 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 at both pointers match, move pointers closer; if not, string is not palindrome.

3. Final Answer:

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

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 both ends moving inward [OK]

Common Mistakes:

• Trying to reverse the string instead of using pointers
• Checking only half the string without comparing both ends
• Ignoring case or non-alphanumeric characters

2. Which of the following JavaScript code snippets correctly initializes two pointers for palindrome checking?

easy

A. let left = s.length - 1, right = 0;

B. let left = 1, right = s.length;

C. let left = s.length, right = 0;

D. let left = 0, right = s.length - 1;

Solution

1. Step 1: Identify correct pointer initialization

   Left pointer should start at 0 (start of string), right pointer at last index (length - 1).

2. Step 2: Check each option

   let left = 0, right = s.length - 1; correctly sets left=0 and right=s.length-1; others have incorrect indices.

3. Final Answer:

   let left = 0, right = s.length - 1; -> Option D

4. Quick Check:

   Start pointers at 0 and length-1 = let left = 0, right = s.length - 1; [OK]

Hint: Start left at 0 and right at length minus one [OK]

Common Mistakes:

• Setting right pointer to s.length (out of bounds)
• Swapping left and right pointers
• Starting left pointer at 1 instead of 0

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

function isPalindrome(s) {
  let left = 0, right = s.length - 1;
  while (left < right) {
    if (s[left] !== s[right]) return false;
    left++;
    right--;
  }
  return true;
}
console.log(isPalindrome('racecar'));
console.log(isPalindrome('hello'));

medium

A. true false

B. false true

C. true true

D. false false

Solution

1. Step 1: Trace isPalindrome('racecar')

   Characters match from both ends moving inward, so returns true.

2. Step 2: Trace isPalindrome('hello')

   Characters at start and end differ ('h' vs 'o'), returns false immediately.

3. Final Answer:

   true false -> Option A

4. Quick Check:

   'racecar' palindrome = true, 'hello' not palindrome = false [OK]

Hint: Check characters from ends; stop at first mismatch [OK]

Common Mistakes:

• Assuming 'hello' is palindrome
• Not returning false immediately on mismatch
• Confusing return values order

4. Find the error in this palindrome check code:

function isPalindrome(s) {
  let left = 0, right = s.length;
  while (left < right) {
    if (s[left] !== s[right]) return false;
    left++;
    right--;
  }
  return true;
}

medium

A. The while condition should be left <= right

B. Left pointer should start at 1, not 0

C. Right pointer should start at s.length - 1, not s.length

D. The function should return false at the end

Solution

1. Step 1: Check pointer initialization

   Right pointer is set to s.length, which is out of string index range (0 to length-1).

2. Step 2: Correct pointer initialization

   Right pointer should be s.length - 1 to point to last character.

3. Final Answer:

   Right pointer should start at s.length - 1, not s.length -> Option C

4. Quick Check:

   Right pointer = length - 1 [OK]

Hint: Right pointer must be last index (length - 1) [OK]

Common Mistakes:

• Using s.length as index causes undefined access
• Changing left pointer start incorrectly
• Incorrect loop condition causing infinite loop

5. Given the string s = 'A man, a plan, a canal: Panama', which modification to the two-pointer palindrome check is needed to correctly identify it as a palindrome?

hard

A. Reverse the string and compare directly

B. Ignore non-alphanumeric characters and compare characters case-insensitively

C. Only check the first half of the string

D. Convert the string to uppercase and compare characters directly

Solution

1. Step 1: Understand the input string

   The string contains spaces, commas, and mixed cases which should be ignored for palindrome check.

2. Step 2: Modify two-pointer approach

   Skip non-alphanumeric characters and compare characters in lowercase to handle case differences.

3. Final Answer:

   Ignore non-alphanumeric characters and compare characters case-insensitively -> Option B

4. Quick Check:

   Clean and lowercase string before comparing [OK]

Hint: Skip symbols and compare lowercase letters only [OK]

Common Mistakes:

• Comparing characters without ignoring symbols
• Not handling uppercase vs lowercase differences
• Assuming direct comparison works on raw string