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

Time Complexity: Valid Palindrome Two Pointer

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a string is a palindrome using two pointers.

How does the time grow as the string gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func isPalindrome(s string) bool {
    left, right := 0, len(s)-1
    for left < right {
        if s[left] != s[right] {
            return false
        }
        left++
        right--
    }
    return true
}

This code checks if a string reads the same forwards and backwards by comparing characters from both ends.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing characters at positions from the start and end moving towards the center.
• How many times: Up to half the length of the string (n/2 times).

How Execution Grows With Input

As the string gets longer, the number of comparisons grows roughly half as fast.

Input Size (n)	Approx. Operations
10	5 comparisons
100	50 comparisons
1000	500 comparisons

Pattern observation: The operations grow linearly with input size, but only half as many comparisons are needed.

Final Time Complexity

Time Complexity: O(n)

This means the time to check grows directly with the length of the string.

Common Mistake

[X] Wrong: "The loop runs n times, so it's O(n²) because of nested comparisons."

[OK] Correct: The loop runs only once through the string, comparing pairs, so it's just O(n), not squared.

Interview Connect

Understanding this simple linear check helps build confidence for string problems and shows you can analyze how efficient your code is.

Self-Check

"What if we ignored non-alphanumeric characters and spaces? How would the time complexity change?"

Practice

1. What does the two-pointer technique check when verifying if a string is a palindrome?

easy

A. It compares characters from the start and end moving towards the center

B. It counts the number of vowels in the string

C. It reverses the string and compares with the original

D. It sorts the characters and checks if sorted equals original

Solution

1. Step 1: Understand two-pointer approach

   The two-pointer technique uses two indexes: one at the start and one at the end of the string.

2. Step 2: Compare characters moving inward

   It compares characters at these pointers and moves them towards the center to check if they match.

3. Final Answer:

   It compares characters from the start and end moving towards the center -> Option A

4. Quick Check:

   Two-pointer = compare start/end chars [OK]

Hint: Check characters from both ends moving inward [OK]

Common Mistakes:

• Thinking it counts vowels instead of comparing characters
• Assuming it reverses the string explicitly
• Confusing sorting with palindrome check

2. Which of the following Go code snippets correctly initializes two pointers for a palindrome check on string s?

easy

A. left := len(s); right := 0

B. left := 1; right := len(s)

C. left := -1; right := len(s) + 1

D. left := 0; right := len(s) - 1

Solution

1. Step 1: Identify valid pointer initialization

   For a string, the first character index is 0 and last is length minus 1.

2. Step 2: Check each option

   left := 0; right := len(s) - 1 correctly sets left to 0 and right to len(s)-1, valid for two-pointer palindrome check.

3. Final Answer:

   left := 0; right := len(s) - 1 -> Option D

4. Quick Check:

   Start=0, End=len(s)-1 [OK]

Hint: Start left at 0 and right at len(s)-1 [OK]

Common Mistakes:

• Starting left at 1 instead of 0
• Setting right to len(s) which is out of range
• Using negative or invalid indices

3. What will be the output of the following Go code?

func isPalindrome(s string) bool {
    left, right := 0, len(s)-1
    for left < right {
        if s[left] != s[right] {
            return false
        }
        left++
        right--
    }
    return true
}

fmt.Println(isPalindrome("racecar"))
fmt.Println(isPalindrome("hello"))

medium

A. true\nfalse

B. false\ntrue

C. true\ntrue

D. false\nfalse

Solution

1. Step 1: Trace "racecar" input

   Characters at start and end match moving inward, so function returns true.

2. Step 2: Trace "hello" input

   Characters 'h' and 'o' differ at first check, so function returns false.

3. Final Answer:

   true\nfalse -> Option A

4. Quick Check:

   "racecar" palindrome = true, "hello" not = false [OK]

Hint: Check first and last chars; mismatch means false [OK]

Common Mistakes:

• Assuming "hello" is palindrome
• Confusing return values order
• Ignoring early return on mismatch

4. Identify the bug in this Go function that checks palindrome:

func isPalindrome(s string) bool {
    left, right := 0, len(s)
    for left < right {
        if s[left] != s[right] {
            return false
        }
        left++
        right--
    }
    return true
}

medium

A. left should start at 1, not 0

B. The loop condition should be left <= right

C. right should be initialized to len(s)-1, not len(s)

D. The function should return false at the end

Solution

1. Step 1: Check pointer initialization

   right is set to len(s), which is out of string index range (0 to len(s)-1).

2. Step 2: Correct initialization

   right should be len(s)-1 to point to last character correctly.

3. Final Answer:

   right should be initialized to len(s)-1, not len(s) -> Option C

4. Quick Check:

   Index out of range error fixed by right = len(s)-1 [OK]

Hint: Set right pointer to len(s)-1 to avoid index error [OK]

Common Mistakes:

• Using len(s) as index causes runtime error
• Changing left pointer start incorrectly
• Misunderstanding loop condition

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 case-insensitively

C. Only compare vowels in the string

D. Convert string to uppercase and compare characters directly

Solution

1. Step 1: Understand input complexity

   The string contains spaces, commas, and mixed cases which should be ignored for palindrome check.

2. Step 2: Modify two-pointer logic

   Skip non-alphanumeric characters and compare characters in lowercase to handle case differences.

3. Final Answer:

   Ignore non-alphanumeric characters and compare case-insensitively -> Option B

4. Quick Check:

   Clean and lowercase string before two-pointer check [OK]

Hint: Skip symbols and compare lowercase chars [OK]

Common Mistakes:

• Not ignoring spaces and punctuation
• Comparing case-sensitively causing false negatives
• Only converting case without skipping symbols