Valid Palindrome Two Pointer in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the left pointer at the start of the string.

DSA Typescript

let left = [1];

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

A-1

B1

Cs.length - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 or negative index.

The left pointer should start at index 0 to begin checking from the start of the string.

2fill in blank

medium

Complete the code to initialize the right pointer at the end of the string.

DSA Typescript

let right = [1];

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

A0

Bs.length

Cs.length - 1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using s.length which is out of bounds.

The right pointer should start at the last index, which is length minus one.

3fill in blank

hard

Fix the error in the while loop condition to continue while left is less than right.

DSA Typescript

while ([1] < right) {

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

Aright

Bleft

Cleft + 1

Dright - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right or incorrect expressions.

The loop should run while the left pointer is less than the right pointer.

4fill in blank

hard

Fill both blanks to skip non-alphanumeric characters from left and right pointers.

DSA Typescript

while (left < right && !s[left].[1](/[a-z0-9]/i)) left++;
while (left < right && !s[right].[2](/[a-z0-9]/i)) right--;

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

AtoLowerCase

BisAlphaNumeric

Cmatch

DtoUpperCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using toLowerCase or toUpperCase instead of checking character type.

Use the match method with a regex to check if characters are alphanumeric.

5fill in blank

hard

Fill all four blanks to compare characters ignoring case and move pointers accordingly.

DSA Typescript

if (s[left].[1]() !== s[right].[2]()) {
  return false;
}
left[3];
right[4];

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

AtoLowerCase

BtoUpperCase

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using different case methods or forgetting to move pointers.

Convert both characters to lowercase to compare ignoring case, then increment left pointer.

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