Valid Palindrome Two Pointer in DSA Javascript - Execution Trace

Concept Flow - Valid Palindrome Two Pointer

Start with two pointers

↓

Check characters at left and right

↓

Move pointers inward

↓

Repeat until left >= right

↓

Return true

Two pointers start at the string ends and move inward, comparing characters. If all pairs match, it's a palindrome.

Execution Sample

DSA Javascript

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

Checks if a string is a palindrome by comparing characters from both ends moving inward.

Execution Table

Step	Operation	left	right	Characters Compared	Match?	Action	Result So Far
1	Initialize pointers	0	6	-	-	Start	Pending
2	Compare s[left] and s[right]	0	6	'r' vs 'r'	Yes	left++, right--	Pending
3	Compare s[left] and s[right]	1	5	'a' vs 'a'	Yes	left++, right--	Pending
4	Compare s[left] and s[right]	2	4	'c' vs 'c'	Yes	left++, right--	Pending
5	Check left < right	3	3	-	-	No (left >= right), return true	True
6	End	3	3	-	-	Return true	Palindrome confirmed

💡 Pointers crossed or met (left >= right), all compared characters matched.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	Final
left	0	1	2	3	3	3
right	6	5	4	3	3	3
Result So Far	Pending	Pending	Pending	Pending	True	True

Key Moments - 3 Insights

Why do we stop when left is equal to or greater than right?

Because when left meets or passes right, all character pairs have been checked. See execution_table row 5 where left and right are both 3, meaning the middle character or crossing point is reached.

What happens if characters at left and right don't match?

The function immediately returns false, as shown in the 'Not Equal?' branch in concept_flow. This stops further checks since palindrome condition fails.

Why do we compare characters before moving pointers?

We must check the current pair before moving inward to ensure all pairs match. Moving pointers first would skip checking the current characters, risking incorrect results.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what are the values of left and right pointers?

A left = 1, right = 5

B left = 2, right = 4

C left = 3, right = 3

D left = 0, right = 6

💡 Hint

Check the 'left' and 'right' columns in execution_table row 3.

At which step does the function decide the string is a palindrome?

A Step 2

B Step 4

C Step 5

D Step 6

💡 Hint

Look at the 'Action' and 'Result So Far' columns where pointers meet or cross.

If the characters at left and right do not match at Step 2, what would happen?

A left and right pointers move inward

B Function returns false immediately

C Function continues checking next characters

D Function returns true immediately

💡 Hint

Refer to concept_flow where 'Not Equal?' leads to 'Return false'.

Concept Snapshot

Valid Palindrome Two Pointer:
- Use two pointers: left at start, right at end
- Compare characters at left and right
- If equal, move pointers inward (left++, right--)
- If not equal, return false immediately
- Stop when left >= right, return true
- Checks palindrome in O(n) time with O(1) space

Full Transcript

This concept uses two pointers starting at the ends of a string. Each step compares the characters at these pointers. If they match, pointers move inward to check the next pair. If any pair does not match, the function returns false immediately. When the pointers meet or cross, it means all pairs matched, so the string is a palindrome. The execution table shows each step with pointer positions, characters compared, and actions taken. The variable tracker records pointer values and result status after each step. Key moments clarify why the loop stops when pointers meet and what happens on mismatch. The visual quiz tests understanding of pointer positions and decision points. This method is efficient and simple for palindrome checking.

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