Why Valid Palindrome Two Pointer in DSA Typescript?

The Big Idea

What if you could check palindromes without flipping the whole word?

The Scenario

Imagine you want to check if a word or sentence reads the same backward as forward, like "madam" or "racecar". Doing this by hand means flipping the word and comparing each letter one by one.

The Problem

Manually reversing and comparing letters is slow and easy to mess up, especially with long sentences or when ignoring spaces and punctuation. It's tiring and error-prone to check each character twice.

The Solution

The two-pointer method uses two markers starting at the beginning and end of the string. They move toward each other, comparing characters only once. This way, you quickly find if the string is a palindrome without extra work.

Before vs After

✗ Before

function isPalindrome(s: string): boolean {
  const reversed = s.split('').reverse().join('');
  return s === reversed;
}

✓ After

function isPalindrome(s: string): boolean {
  let left = 0;
  let right = s.length - 1;
  while (left < right) {
    if (s[left] !== s[right]) return false;
    left++;
    right--;
  }
  return true;
}

What It Enables

This method lets you quickly and efficiently check palindromes even in long texts or complex strings.

Real Life Example

Checking if a username or password is a palindrome to enforce security rules or for fun features in apps.

Key Takeaways

Manual reversal is slow and error-prone.

Two pointers compare characters from both ends efficiently.

It works well for long strings and ignores extra work.

Practice

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

easy

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

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

C. Use two pointers starting at both ends and move towards the center comparing characters

D. Sort the characters and check if the sorted string equals the original

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 at both pointers match, move pointers closer; otherwise, it's not a palindrome.

3. Final Answer:

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

4. Quick Check:

   Two-pointer palindrome check = Use two pointers starting at both ends and move towards the center comparing characters [OK]

Hint: Compare characters from both ends moving inward [OK]

Common Mistakes:

• Reversing string instead of using pointers
• Checking only half the string without pointers
• Ignoring case or non-alphanumeric characters

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

easy

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

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

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

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

Solution

1. Step 1: Identify pointer positions

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

2. Step 2: Check code correctness

   let left = 0; let right = s.length - 1; correctly sets left = 0 and right = s.length - 1, valid indices for string.

3. Final Answer:

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

4. Quick Check:

   Pointers start at 0 and length-1 = let left = 0; let 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)
• Starting left pointer at 1 instead of 0
• Confusing length with last index

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

function isPalindrome(s: string): boolean {
  let left = 0;
  let 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 'racecar' input

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

2. Step 2: Trace 'hello' input

   Characters 'h' and 'o' differ at ends, returns false immediately.

3. Final Answer:

   true false -> Option A

4. Quick Check:

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

Hint: Check characters from ends; mismatch means false [OK]

Common Mistakes:

• Assuming 'hello' is palindrome
• Returning true without checking all pairs
• Off-by-one errors in pointer movement

4. Identify the bug in this TypeScript palindrome checker:

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

medium

A. left pointer should start at 1

B. right pointer initialized to s.length instead of s.length - 1

C. Comparison should be s[left] === s[right]

D. Loop condition should be left <= right

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: Understand impact of bug

   Accessing s[s.length] causes undefined comparison or runtime error.

3. Final Answer:

   right pointer initialized to s.length instead of s.length - 1 -> Option B

4. Quick Check:

   Right pointer must be last index = length-1 [OK]

Hint: Right pointer must be last index, not length [OK]

Common Mistakes:

• Using s.length as index causes out-of-bounds
• Changing loop condition incorrectly
• Confusing equality and inequality in comparison

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

hard

A. Convert string to uppercase and compare all characters

B. Reverse the string and compare directly

C. Only compare characters at even indices

D. Ignore case and skip non-alphanumeric characters during comparison

Solution

1. Step 1: Understand input complexity

   String contains spaces, commas, and mixed case letters which must be ignored for palindrome check.

2. Step 2: Modify two-pointer logic

   Skip non-alphanumeric characters and compare characters case-insensitively while moving pointers inward.

3. Final Answer:

   Ignore case and skip non-alphanumeric characters during comparison -> Option D

4. Quick Check:

   Skip non-letters and ignore case for valid palindrome [OK]

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

Common Mistakes:

• Comparing characters without ignoring punctuation
• Not converting characters to same case
• Assuming direct comparison works on raw string