Valid Palindrome Two Pointer in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Two Pointer Palindrome Check

What is the output of the following code when checking if the string is a valid palindrome ignoring non-alphanumeric characters and case?

DSA Python

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
        if s[left].lower() != s[right].lower():
            return False
        left += 1
        right -= 1
    return True

print(is_palindrome("A man, a plan, a canal: Panama"))

A False

B SyntaxError

C TypeError

D True

💡 Hint

Check how non-alphanumeric characters are skipped and case is ignored.

The code uses two pointers moving inward, skipping non-alphanumeric characters and comparing characters case-insensitively. The input string is a well-known palindrome phrase ignoring punctuation and case, so the function returns True.

❓ Predict Output

intermediate

Output for Non-Palindrome String

What will the function return for the input string "race a car"?

DSA Python

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
        if s[left].lower() != s[right].lower():
            return False
        left += 1
        right -= 1
    return True

print(is_palindrome("race a car"))

A False

B True

C IndexError

D ValueError

💡 Hint

Check the characters compared at the first mismatch.

The function returns False because "race a car" is not a palindrome when ignoring spaces and case. The characters compared at the mismatch cause the function to return False.

🔧 Debug

advanced

Identify the bug in palindrome check

What error will this code produce when run with input "No lemon, no melon"?

DSA Python

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left <= right:
        if not s[left].isalnum():
            left += 1
        if not s[right].isalnum():
            right -= 1
        if s[left].lower() != s[right].lower():
            return False
        left += 1
        right -= 1
    return True

print(is_palindrome("No lemon, no melon"))

A IndexError

B True

C False

D SyntaxError

💡 Hint

Check how pointers move when skipping characters.

The code increments left and right pointers without using 'while' loops to skip non-alphanumeric characters properly. This can cause pointers to move past each other and cause IndexError when accessing s[left] or s[right].

❓ Predict Output

advanced

Output of palindrome check with empty string

What is the output of the palindrome check function when the input string is empty?

DSA Python

def is_palindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
        if s[left].lower() != s[right].lower():
            return False
        left += 1
        right -= 1
    return True

print(is_palindrome("") )

A IndexError

B False

C True

D TypeError

💡 Hint

Consider the meaning of an empty string as a palindrome.

An empty string is considered a valid palindrome. The function returns True immediately because the while loop condition is false initially.

🧠 Conceptual

expert

Why use two pointers for palindrome check?

Which of the following best explains why the two-pointer technique is efficient for checking if a string is a palindrome ignoring non-alphanumeric characters and case?

A It converts the string to a list and reverses it, then compares both lists for equality.

B It allows checking characters from both ends simultaneously, skipping irrelevant characters, reducing time complexity to O(n).

C It uses recursion to compare characters, which uses less memory than iteration.

D It sorts the string first, then compares characters, which is faster than other methods.

💡 Hint

Think about how skipping characters and comparing from both ends helps.

The two-pointer technique moves inward from both ends, skipping non-alphanumeric characters and comparing characters in one pass, making it efficient with O(n) time complexity.