Valid Palindrome
Detailed guide and Python implementation for the 'Valid Palindrome' problem.
1. Concept Overview
The 'Valid Palindrome' problem is a key challenge in the Two Pointers section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Valid Palindrome.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Valid Palindrome carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string s, return True if it is a palindrome, or False otherwise.
Write a function isPalindrome(s: str) -> bool.
- •1 <= len(s) <= 2 * 10^5
- •s consists only of printable ASCII characters
Examples
s = "A man, a plan, a canal: Panama"
True
After removing non-alphanumeric characters and converting to lowercase: "amanaplanacanalpanama", which is a palindrome.
s = "race a car"
False
After processing: "raceacar" is not a palindrome.
s = " "
True
After removing non-alphanumeric characters, s is an empty string. An empty string is a palindrome by definition.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def is_palindrome_opt(s):
l, r = 0, len(s) - 1
while l < r:
while l < r and not s[l].isalnum():
l += 1
while r > l and not s[r].isalnum():
r -= 1
if s[l].lower() != s[r].lower():
return False
l, r = l + 1, r - 1
return TrueBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def is_palindrome_brute(s):
new_s = ""
for c in s:
if c.isalnum():
new_s += c.lower()
return new_s == new_s[::-1]Algorithm Pattern Checklist
When dealing with Two Pointers data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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