Palindrome Detection in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a palindrome?

A palindrome is a sequence of characters that reads the same forward and backward, like "madam" or "racecar".

beginner

How can you check if a string is a palindrome using two pointers?

Use two pointers: one at the start and one at the end of the string. Move both pointers towards the center, comparing characters. If all pairs match, it's a palindrome.

intermediate

Why is ignoring case and non-alphanumeric characters important in palindrome detection?

Ignoring case and non-alphanumeric characters helps detect palindromes in phrases or sentences, like "A man, a plan, a canal: Panama", which are palindromes when cleaned.

intermediate

What is the time complexity of checking a palindrome by comparing characters from both ends?

The time complexity is O(n), where n is the length of the string, because each character is checked at most once.

beginner

Show a simple Go function signature for palindrome detection.

func isPalindrome(s string) bool

Which of the following strings is a palindrome?

A hello

B level

C world

D golang

"level" reads the same forward and backward, so it is a palindrome.

What is the first step in checking if a string is a palindrome?

A Compare characters from the middle outwards

B Reverse the string and compare

C Use two pointers at start and end to compare characters

D Count the number of vowels

Using two pointers at the start and end to compare characters is a common and efficient method.

What should you do with spaces and punctuation when checking palindromes in sentences?

A Ignore them

B Count them as characters

C Replace them with numbers

D Double them

Ignoring spaces and punctuation helps detect palindromes in sentences correctly.

What is the time complexity of palindrome detection using two pointers?

A O(log n)

B O(1)

C O(n^2)

D O(n)

Each character is checked once, so the time complexity is linear, O(n).

In Go, what type should the input string have for palindrome detection?

A string

B int

C bool

D float64

The input should be a string type to check if it is a palindrome.

Explain how to check if a string is a palindrome using the two-pointer technique.

Think about comparing characters from both ends moving inward.

Describe why preprocessing a string (like ignoring case and non-alphanumeric characters) is useful in palindrome detection.

Consider how phrases with spaces and punctuation can still be palindromes.

Practice

1. What does a palindrome string mean in simple terms?

easy

A. A string with all unique characters

B. A string that contains only vowels

C. A string that reads the same forwards and backwards

D. A string that ends with a special character

Solution

1. Step 1: Understand palindrome definition

   A palindrome is a word or phrase that reads the same from left to right and right to left.

2. Step 2: Match options with definition

   A string that reads the same forwards and backwards matches this definition exactly, others describe different string properties.

3. Final Answer:

   A string that reads the same forwards and backwards -> Option C

4. Quick Check:

   Palindrome = Same forwards and backwards [OK]

Hint: Check if string equals its reverse [OK]

Common Mistakes:

• Confusing palindrome with unique characters
• Thinking palindrome means only vowels
• Assuming palindrome ends with special character

2. Which Go code snippet correctly checks if a string s is a palindrome?

easy

A. for i := 0; i < len(s)/2; i++ { if s[i] != s[len(s)-1-i] { return false } } return true

B. for i := 0; i < len(s); i++ { if s[i] == s[i+1] { return true } } return false

C. for i := 0; i < len(s)/2; i++ { if s[i] == s[len(s)-1-i] { return false } } return true

D. for i := 0; i < len(s); i++ { if s[i] != s[i] { return false } } return true

Solution

1. Step 1: Analyze palindrome check logic

   We compare characters from start and end moving towards center. If any pair differs, return false.

2. Step 2: Match code with logic

   for i := 0; i < len(s)/2; i++ { if s[i] != s[len(s)-1-i] { return false } } return true correctly compares s[i] with s[len(s)-1-i] for half the string length and returns false if mismatch found.

3. Final Answer:

   for i := 0; i < len(s)/2; i++ { if s[i] != s[len(s)-1-i] { return false } } return true -> Option A

4. Quick Check:

   Compare pairs from ends = for i := 0; i < len(s)/2; i++ { if s[i] != s[len(s)-1-i] { return false } } return true [OK]

Hint: Compare characters from ends to middle [OK]

Common Mistakes:

• Checking only adjacent characters
• Returning false when characters match
• Comparing character with itself incorrectly

3. What is the output of this Go code snippet?

func isPalindrome(s string) bool {
    for i := 0; i < len(s)/2; i++ {
        if s[i] != s[len(s)-1-i] {
            return false
        }
    }
    return true
}

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

medium

A. false false

B. false true

C. true true

D. true false

Solution

1. Step 1: Check "level" palindrome

   "level" reads same forwards and backwards, so function returns true.

2. Step 2: Check "hello" palindrome

   "hello" reversed is "olleh", different, so function returns false.

3. Final Answer:

   true false -> Option D

4. Quick Check:

   "level"=true, "hello"=false [OK]

Hint: Test string and its reverse equality [OK]

Common Mistakes:

• Assuming all strings are palindrome
• Mixing output order
• Ignoring case sensitivity

4. Find the error in this Go function that checks palindrome:

func isPalindrome(s string) bool {
    for i := 0; i <= len(s)/2; i++ {
        if s[i] != s[len(s)-1-i] {
            return false
        }
    }
    return true
}

medium

A. The loop condition should be 'i < len(s)/2' to avoid index out of range

B. The comparison should be 's[i] == s[len(s)-1-i]' to return true early

C. The function should return false at the end instead of true

D. The loop should start from 'i := 1' instead of 0

Solution

1. Step 1: Analyze loop boundary

   The loop uses 'i <= len(s)/2', which causes 'i' to reach len(s)/2 inclusive, possibly exceeding index for odd length strings.

2. Step 2: Correct loop condition

   Changing to 'i < len(s)/2' ensures no index out of range error and proper comparison only up to middle.

3. Final Answer:

   The loop condition should be 'i < len(s)/2' to avoid index out of range -> Option A

4. Quick Check:

   Loop boundary off by one causes error [OK]

Hint: Use '<' not '<=' for half-length loops [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Returning wrong boolean at end
• Starting loop from 1 instead of 0

5. You want to detect if a string is a palindrome ignoring case and spaces. Which approach is best in Go?

hard

A. Check palindrome directly without changes

B. Remove spaces, convert to lowercase, then check palindrome normally

C. Only convert to lowercase, ignore spaces

D. Reverse string and compare without any changes

Solution

1. Step 1: Understand problem requirements

   We must ignore spaces and case differences when checking palindrome.

2. Step 2: Prepare string correctly

   Removing spaces and converting all letters to lowercase normalizes the string for accurate palindrome check.

3. Step 3: Check palindrome

   After normalization, use standard palindrome check comparing characters from ends.

4. Final Answer:

   Remove spaces, convert to lowercase, then check palindrome normally -> Option B

5. Quick Check:

   Normalize string before check = Remove spaces, convert to lowercase, then check palindrome normally [OK]

Hint: Clean string first: lowercase + remove spaces [OK]

Common Mistakes:

• Ignoring spaces in palindrome check
• Not converting to lowercase
• Comparing raw strings directly