Valid Palindrome Two Pointer in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

int left = [1];

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

As.length()

B1

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 skips the first character.

Using s.length() causes index out of bounds.

Using -1 is invalid 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 Java

int right = [1];

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

As.length() - 1

B0

Cs.length()

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using s.length() causes index out of bounds.

Using 0 starts at the beginning, not the end.

Using -1 is invalid index.

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

3fill in blank

hard

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

DSA Java

while ([1] < right) {

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

A0

Bright

Cs.length()

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right is always false.

Using s.length() < right is incorrect comparison.

Using 0 < right ignores left pointer.

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 left and right pointers.

DSA Java

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

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

AisLetterOrDigit

BisWhitespace

CisDigit

DisLetter

Attempts:

3 left

💡 Hint

Common Mistakes

Using isWhitespace skips spaces only, not all non-alphanumeric.

Using isDigit skips only digits, missing letters.

Using isLetter skips only letters, missing digits.

We skip characters that are not letters or digits on both sides.

5fill in blank

hard

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

DSA Java

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

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

AtoLowerCase

BtoUpperCase

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using different case conversions on left and right causes mismatch.

Not moving pointers causes infinite loop.

Using decrement on left pointer is wrong direction.

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

Practice

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

easy

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

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

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

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

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 B

4. Quick Check:

   Two-pointer technique = Compare ends inward [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 Java code snippets correctly initializes two pointers for palindrome checking on string s?

easy

A. int left = 0, right = s.length() - 1;

B. int left = 0, right = s.length();

C. int left = 1, right = s.length() - 1;

D. int left = 0, right = s.length() + 1;

Solution

1. Step 1: Identify valid pointer initialization

   String indices start at 0 and end at length - 1, so right pointer must be s.length() - 1.

2. Step 2: Check each option

   int left = 0, right = s.length() - 1; correctly sets left to 0 and right to s.length() - 1; others are off by one or invalid.

3. Final Answer:

   int left = 0, right = s.length() - 1; -> Option A

4. Quick Check:

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

Hint: Right pointer = string length minus one [OK]

Common Mistakes:

• Setting right pointer to s.length() causing out-of-bounds
• Starting left pointer at 1 instead of 0
• Using length + 1 causing errors

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

String s = "A man, a plan, a canal: Panama";
boolean isPalindrome = true;
int left = 0, right = s.length() - 1;
while (left < right) {
  char leftChar = Character.toLowerCase(s.charAt(left));
  char rightChar = Character.toLowerCase(s.charAt(right));
  if (!Character.isLetterOrDigit(leftChar)) {
    left++;
  } else if (!Character.isLetterOrDigit(rightChar)) {
    right--;
  } else {
    if (leftChar != rightChar) {
      isPalindrome = false;
      break;
    }
    left++;
    right--;
  }
}
System.out.println(isPalindrome);

medium

A. Compilation error

B. false

C. true

D. Runtime exception

Solution

1. Step 1: Understand the code logic

   The code skips non-alphanumeric characters and compares lowercase letters from both ends moving inward.

2. Step 2: Trace the string "A man, a plan, a canal: Panama"

   Ignoring spaces and punctuation, the cleaned string is "amanaplanacanalpanama" which reads the same forward and backward.

3. Final Answer:

   true -> Option C

4. Quick Check:

   Ignoring non-alphanumeric, palindrome check = true [OK]

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

Common Mistakes:

• Not skipping non-alphanumeric characters
• Not converting characters to lowercase
• Breaking loop too early without full check

4. Identify the bug in this Java code snippet for palindrome checking:

String s = "racecar";
int left = 0, right = s.length() - 1;
boolean isPalindrome = true;
while (left <= right) {
  if (s.charAt(left) != s.charAt(right)) {
    isPalindrome = false;
    break;
  }
  left++;
  right--;
}
System.out.println(isPalindrome);

medium

A. The code does not skip non-alphanumeric characters

B. The pointers should start at 1 and length - 2

C. The loop condition should be left < right, not left <= right

D. The code does not convert characters to lowercase

Solution

1. Step 1: Analyze loop condition

   Using left <= right causes the middle character to be compared with itself unnecessarily, but it does not cause wrong result here.

2. Step 2: Check correctness of condition

   The main issue is that the code does not convert characters to lowercase, so it fails for case-insensitive palindrome checks.

3. Final Answer:

   The code does not convert characters to lowercase -> Option D

4. Quick Check:

   Case sensitivity causes incorrect palindrome detection [OK]

Hint: Convert characters to lowercase to handle case sensitivity [OK]

Common Mistakes:

• Using <= causes unnecessary middle char check
• Ignoring case sensitivity
• Not handling non-alphanumeric characters

5. Given the string s = "No lemon, no melon", which modification to the two-pointer palindrome check is necessary to correctly identify it as a palindrome?

hard

A. Convert characters to lowercase and skip non-alphanumeric characters during comparison

B. Only compare characters without any conversion or skipping

C. Reverse the string and compare directly without pointers

D. Use a stack to store characters and compare with the original

Solution

1. Step 1: Understand the string content

   The string contains spaces and uppercase letters which should be ignored or normalized for palindrome check.

2. Step 2: Modify two-pointer approach

   Convert characters to lowercase and skip non-alphanumeric characters to compare only valid letters.

3. Final Answer:

   Convert characters to lowercase and skip non-alphanumeric characters during comparison -> Option A

4. Quick Check:

   Normalize case and skip invalid chars for correct palindrome [OK]

Hint: Normalize case and ignore spaces/punctuation [OK]

Common Mistakes:

• Not ignoring spaces and punctuation
• Not converting uppercase to lowercase
• Assuming direct comparison without cleaning works