Valid Palindrome Two Pointer in DSA Java - Time & Space Complexity

Time Complexity: Valid Palindrome Two Pointer

O(n)

Understanding Time Complexity

We want to know how the time needed to check if a string is a palindrome grows as the string gets longer.

Specifically, how many steps does the two-pointer method take as input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public boolean isPalindrome(String s) {
    int left = 0, right = s.length() - 1;
    while (left < right) {
        while (left < right && !Character.isLetterOrDigit(s.charAt(left))) left++;
        while (left < right && !Character.isLetterOrDigit(s.charAt(right))) right--;
        if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) return false;
        left++;
        right--;
    }
    return true;
}
    

This code checks if a string is a palindrome by moving two pointers from both ends towards the center, skipping non-alphanumeric characters.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves two pointers from the start and end towards the middle.
• How many times: Each character is visited at most once by either pointer, so up to n times total, where n is the string length.

How Execution Grows With Input

As the string length grows, the number of steps grows roughly in direct proportion.

Input Size (n)	Approx. Operations
10	About 10 checks
100	About 100 checks
1000	About 1000 checks

Pattern observation: The steps increase linearly as the input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to check grows in a straight line with the string length.

Common Mistake

[X] Wrong: "Because there are nested while loops, the time complexity is O(n²)."

[OK] Correct: Each character is visited only once by either pointer, so the total steps add up to n, not n times n.

Interview Connect

Understanding this linear time approach shows you can efficiently process strings with two pointers, a common and useful skill in coding challenges.

Self-Check

"What if we changed the code to check only letters and ignored digits? How would the time complexity change?"

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