Valid Palindrome Two Pointer in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

let left = [1];

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

A0

Bs.length

Cs.length - 1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 skips the first character.

Using s.length or s.length - 1 for left pointer start causes out of range or wrong start.

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 Javascript

let right = [1];

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

As.length

Bs.length - 1

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using s.length causes index out of range.

Starting right pointer at 0 means only checking the first character.

The right pointer should start at the last character index, which is s.length - 1.

3fill in blank

hard

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

DSA Javascript

while ([1] < right) {

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

Aleft + 1

Bright

Cleft

Dright - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right is always false, loop never runs.

Using left + 1 < right causes skipping first comparison.

The loop should run while left pointer is less than right pointer to compare characters.

4fill in blank

hard

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

DSA Javascript

while (left < right && !s[left].match(/[1]/)) left++;
while (left < right && !s[right].match(/[2]/)) right--;

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

A[a-zA-Z0-9]

B[^a-zA-Z0-9]

C[0-9]

D[a-z]

Attempts:

3 left

💡 Hint

Common Mistakes

Using [^a-zA-Z0-9] matches non-alphanumeric, which is opposite of what we want here.

Using only [a-z] misses digits and uppercase letters.

We want to skip characters that are NOT alphanumeric, so we check if current character does NOT match alphanumeric pattern.

5fill in blank

hard

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

DSA Javascript

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

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

A!==

B===

C--

D++

Attempts:

3 left

💡 Hint

Common Mistakes

Using === instead of !== causes wrong palindrome check.

Incrementing right pointer or decrementing left pointer reverses logic.

If characters are not equal ignoring case, return false. Otherwise, move left pointer forward and right pointer backward.

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