Why Valid Palindrome Two Pointer in DSA Go?

The Big Idea

What if you could check palindromes instantly without flipping the whole word?

The Scenario

Imagine you want to check if a word or sentence reads the same backward as forward, like "madam" or "racecar". Doing this by hand means flipping the word and comparing each letter one by one.

The Problem

Manually reversing and comparing each letter is slow and easy to mess up, especially with long sentences or when ignoring spaces and punctuation. It's tiring and error-prone.

The Solution

The two-pointer method uses two markers starting at the beginning and end of the string. They move towards each other, comparing letters quickly and skipping unwanted characters. This makes checking palindromes fast and simple.

Before vs After

✗ Before

func isPalindrome(s string) bool {
    cleaned := ""
    reversed := ""
    for i := 0; i < len(s); i++ {
        c := s[i]
        if (c >= '0' && c <= '9') || (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') {
            if c >= 'A' && c <= 'Z' {
                c += 32
            }
            cleaned += string(c)
        }
    }
    for i := len(cleaned) - 1; i >= 0; i-- {
        reversed += string(cleaned[i])
    }
    return cleaned == reversed
}

✓ After

func isPalindrome(s string) bool {
    left, right := 0, len(s)-1
    for left < right {
        for left < right && !((s[left] >= 'a' && s[left] <= 'z') || (s[left] >= 'A' && s[left] <= 'Z') || (s[left] >= '0' && s[left] <= '9')) {
            left++
        }
        for left < right && !((s[right] >= 'a' && s[right] <= 'z') || (s[right] >= 'A' && s[right] <= 'Z') || (s[right] >= '0' && s[right] <= '9')) {
            right--
        }
        if left < right {
            l := s[left]
            if l >= 'A' && l <= 'Z' {
                l += 32
            }
            r := s[right]
            if r >= 'A' && r <= 'Z' {
                r += 32
            }
            if l != r {
                return false
            }
            left++
            right--
        }
    }
    return true
}

What It Enables

This method lets you quickly and efficiently check if any string is a palindrome, even ignoring spaces and punctuation.

Real Life Example

Checking if a phrase like "A man, a plan, a canal: Panama" is a palindrome without manually cleaning and reversing it saves time and avoids mistakes.

Key Takeaways

Manual reversal is slow and error-prone.

Two pointers compare characters from both ends efficiently.

Works well even with spaces and punctuation when combined with skipping logic.

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