Valid Palindrome Two Pointer in DSA Javascript - Build from Scratch

Valid Palindrome Using Two Pointers

📖 Scenario: Imagine you are building a simple text checker that tells if a word or phrase reads the same forwards and backwards, ignoring spaces, punctuation, and letter case. This is useful in games, puzzles, or text analysis.

🎯 Goal: You will create a program that checks if a given string is a palindrome using the two-pointer technique. This means comparing characters from the start and end moving towards the center.

📋 What You'll Learn

Create a string variable named inputString with the exact value "A man, a plan, a canal: Panama".

Create a variable named left and set it to 0.

Create a variable named right and set it to inputString.length - 1.

Use a while loop that runs while left < right.

Inside the loop, skip non-alphanumeric characters by moving left forward and right backward.

Compare characters at left and right ignoring case.

If characters do not match, set a variable isPalindrome to false and break the loop.

If loop finishes without mismatch, isPalindrome should be true.

Print isPalindrome.

💡 Why This Matters

🌍 Real World

Checking palindromes is useful in text analysis, puzzles, and validating special strings like IDs or codes.

💼 Career

Understanding two-pointer techniques helps in coding interviews and solving string manipulation problems efficiently.

1

Create the input string

Create a string variable called inputString and set it to the exact value "A man, a plan, a canal: Panama".

DSA Javascript

# Create the input string
# Your code here

Hint

Use const inputString = "A man, a plan, a canal: Panama"; to create the string.

2

Set up two pointers

Create a variable called left and set it to 0. Create another variable called right and set it to inputString.length - 1.

DSA Javascript

const inputString = "A man, a plan, a canal: Panama";
# Create variables left and right pointers
# Your code here

Hint

Use let left = 0; and let right = inputString.length - 1;.

3

Implement the two-pointer palindrome check

Create a variable called isPalindrome and set it to true. Use a while loop that runs while left < right. Inside the loop, skip non-alphanumeric characters by moving left forward and right backward. Compare characters at left and right ignoring case. If characters do not match, set isPalindrome to false and break the loop. Otherwise, move left forward and right backward.

DSA Javascript

const inputString = "A man, a plan, a canal: Panama";
let left = 0;
let right = inputString.length - 1;
# Implement the two-pointer palindrome check
# Your code here

Hint

Use while loops to skip non-alphanumeric characters and compare characters ignoring case.

4

Print the result

Print the value of the variable isPalindrome.

DSA Javascript

const inputString = "A man, a plan, a canal: Panama";
let left = 0;
let right = inputString.length - 1;
let isPalindrome = true;

while (left < right) {
    while (left < right && !/[a-zA-Z0-9]/.test(inputString[left])) {
        left++;
    }
    while (left < right && !/[a-zA-Z0-9]/.test(inputString[right])) {
        right--;
    }
    if (inputString[left].toLowerCase() !== inputString[right].toLowerCase()) {
        isPalindrome = false;
        break;
    }
    left++;
    right--;
}
# Print the result
# Your code here

Hint

Use console.log(isPalindrome); to print the result.

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