🏷️ Category: Strings | 🎯 Difficulty:
Easy| 🔥 Frequency: ★★★★★
📋 Problem Statement
Given a string, return True if it reads the same forwards and backwards, else False. Ignore case and non-alphanumeric characters.
📥 Input → "racecar" 📤 Output → True
📥 Input → "A man a plan a canal Panama" 📤 Output → True
📥 Input → "hello" 📤 Output → False
🧠 What the Interviewer is Testing
| Concept | Why It Matters |
|---|---|
| String cleaning | Real-world input handling |
| isalnum() method | Python string methods |
| Comparison with reverse | Pythonic thinking |
💡 Approach
- Clean the string — lowercase + keep only alphanumeric characters
- Compare cleaned string with its reverse
"A man a plan a canal Panama"
→ clean → "amanaplanacanalpanama"
→ reverse → "amanaplanacanalpanama"
→ equal? ✅ True
✅ Optimal Solution
def is_palindrome(s: str) -> bool:
# Step 1: Clean — lowercase + alphanumeric only
cleaned = ''.join(c.lower() for c in s if c.isalnum())
# Step 2: Compare with its reverse
return cleaned == cleaned[::-1]
# ── Test Cases ─────────────────────────────────
print(is_palindrome("racecar")) # ✅ True
print(is_palindrome("A man a plan a canal Panama")) # ✅ True
print(is_palindrome("hello")) # ✅ False
print(is_palindrome("")) # ✅ True (empty is palindrome)
🔄 Two Pointer Alternative
def is_palindrome_two_ptr(s: str) -> bool:
cleaned = ''.join(c.lower() for c in s if c.isalnum())
left, right = 0, len(cleaned) - 1
while left < right:
if cleaned[left] != cleaned[right]:
return False
left += 1
right -= 1
return True
💬 Pro tip: Mention the two-pointer approach as an alternative — it shows deeper understanding and O(1) space awareness.
📊 Complexity Analysis
┌─────────────┬────────────────────────────────────┐
│ Time O(n) │ Clean + reverse both are O(n) │
│ Space O(n) │ Cleaned string stored in memory │
└─────────────┴────────────────────────────────────┘
🎤 How to Explain in the Interview
“First I clean the string — convert to lowercase and keep only alphanumeric characters. Then I compare it with its reverse. If they’re equal, it’s a palindrome. I can also use a two-pointer approach which avoids creating the reversed string.”
🔁 Common Follow-up Questions
Q1 → What about numbers like 121?
def is_palindrome_num(n: int) -> bool:
s = str(n)
return s == s[::-1]
Q2 → Without converting to string?
# Math approach — reverse the number
def is_palindrome_math(n: int) -> bool:
if n < 0:
return False
original, reversed_n = n, 0
while n > 0:
reversed_n = reversed_n * 10 + n % 10
n //= 10
return original == reversed_n
📁 Set 1 of 4 — Python Interview Prep Series — devprepguide.com