Strictly Palindromic Number requires checking if n is palindromic in every base from 2 to n - 2. Using two-pointer scanning, you verify each representation quickly and track invariants to avoid unnecessary conversions. This approach efficiently identifies failures early and returns false as soon as a non-palindromic base is found.
Problem Statement
A number n is strictly palindromic if for every base b between 2 and n - 2, the representation of n in base b reads the same forwards and backwards. Your task is to check this property efficiently.
Given an integer n, return true if it is strictly palindromic and false otherwise. A string representation is palindromic if the first and last characters match and this property holds recursively inward, making two-pointer scanning a natural pattern to use.
Examples
Example 1
Input: n = 9
Output: false
In base 2: 9 = 1001 (base 2), which is palindromic. In base 3: 9 = 100 (base 3), which is not palindromic. Therefore, 9 is not strictly palindromic so we return false. Note that in bases 4, 5, 6, and 7, n = 9 is also not palindromic.
Example 2
Input: n = 4
Output: false
We only consider base 2: 4 = 100 (base 2), which is not palindromic. Therefore, we return false.
Constraints
- 4 <= n <= 105
Solution Approach
Convert Number to Base
For each base b from 2 to n - 2, convert n to a string in that base. This allows you to inspect the digits directly and prepare for palindrome verification using two pointers.
Two-Pointer Palindrome Check
Use a left and right pointer to scan the base-b string representation of n. Move the pointers inward and compare digits. If any mismatch occurs, immediately return false, as n is not strictly palindromic.
Early Exit and Invariant Tracking
Track invariants like the length and first/last digit matches to optimize scanning. Stop checking further bases once a non-palindromic representation is found, reducing unnecessary computation.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(n log n) due to converting n into multiple bases and scanning each string. Space complexity is O(log n) for storing each base representation during the two-pointer check.
What Interviewers Usually Probe
- Asks how you would verify palindromic property in multiple bases efficiently.
- Hints at using two-pointer scanning instead of reversing strings for optimization.
- Questions about early exit conditions and invariant tracking to avoid extra work.
Common Pitfalls or Variants
Common pitfalls
- Assuming n can be strictly palindromic for n >= 4, which is always false in practice.
- Not stopping early when a base fails, leading to unnecessary computation.
- Using full string reversal for palindrome check instead of two-pointer comparison.
Follow-up variants
- Check if a number is palindromic only in prime bases between 2 and n - 2.
- Return all bases where n is not palindromic rather than just a boolean.
- Verify strictly palindromic property for a list of numbers efficiently using batch processing.
How GhostInterview Helps
- GhostInterview guides you to immediately identify the two-pointer pattern for base conversions.
- It highlights early exit strategies and invariant tracking to avoid redundant scanning.
- It flags failure modes specific to strictly palindromic numbers, ensuring correct boolean results fast.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is a strictly palindromic number?
A strictly palindromic number reads the same forwards and backwards in every base from 2 to n - 2.
Why use two-pointer scanning for this problem?
Two-pointer scanning allows checking palindrome property in linear time without reversing strings, matching the problem pattern efficiently.
Can any number greater than 4 be strictly palindromic?
No, any integer n >= 4 is never strictly palindromic because in base n - 2, the representation is always 12, which is not a palindrome.
How do I optimize checking all bases?
Track invariants and stop as soon as a base produces a non-palindromic representation to reduce computation.
What should I focus on during interview for this problem?
Emphasize the two-pointer scanning with invariant tracking and clearly explain early exit reasoning for strictly palindromic checks.
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