How to Solve Palindrome Number on LeetCode
The Palindrome Number problem is a common LeetCode and screening question, testing your handling of integer operations and edge case thinking.
Problem Statement
Given an integer x, return true if x is palindrome integer. An integer is a palindrome when it reads the same backward as forward.
Example:
Input: x = 121
Output: true
Why This Problem Matters for Interviews
This problem:
- Assesses ability to manipulate numbers and edge cases
- Encourages discussion of both string and math approaches
- Reveals understanding of negative numbers, overflows, and reversals
Approaches to Palindrome Number
1. String Conversion
Compare string and its reverse.
Time Complexity: O(log n)
Space Complexity: O(log n)
def isPalindrome(x: int) -> bool:
return str(x) == str(x)[::-1]
2. Math Approach (No String)
Reverse half of the number and compare.
Time Complexity: O(log n)
Space Complexity: O(1)
def isPalindrome(x: int) -> bool:
if x < 0 or (x % 10 == 0 and x != 0):
return False
rev = 0
while x > rev:
rev = rev * 10 + x % 10
x //= 10
return x == rev or x == rev // 10
Key Interview Talking Points
- Negative numbers are never palindromes
- Avoid string conversion if asked
- When overflows matter (very large numbers)
- Edge cases: single digit, ending in zero
Big O Complexity Recap
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| String | O(log n) | O(log n) |
| Math | O(log n) | O(1) |
Pro Interview Tips
- Explain how to avoid string conversion.
- Discuss negative numbers and why they're not palindromes.
- Practice with test cases