EASY NC#10 Blind #56 Two Pointers
125. Valid Palindrome
📖 Problem
Given a string s, return true if it is a palindrome, or false otherwise. 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.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •In-place array updates
- •Sorted array traversal
- •Boundary condition checks
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Two Pointers reasoning pattern
💡 Approach
- → Use two pointers from both ends
- → Move pointers inward while matching
- → Skip non-alphanumeric characters
- → Ignore case
- → Time: O(n), Space: O(1)
🛠️ Hints & Pitfalls
Hints
- •Use two pointers from both ends
- •Move pointers inward while matching
- •Skip non-alphanumeric characters
Common Pitfalls
- •Ignore case
- •Time: O(n), Space: O(1)
🧪 Test Cases
Hidden tests on submit: 5
Test Case 1
Not run Input:
isPalindrome('A man, a plan, a canal: Panama'); Expected:
true Test Case 2
Not run Input:
isPalindrome(''); Expected:
true Test Case 3
Not run Input:
isPalindrome('a'); Expected:
true 📝 Code Editor
📤 Output