LeetCode 2396 - Strictly Palindromic Number

LeetCode Problem 2396

Difficulty: 🟡 Medium
Topics: Math, Two Pointers, Brainteaser

Solution

Problem Understanding

The problem asks whether a given integer n is strictly palindromic. A number is strictly palindromic if, for every base b between 2 and n - 2 (inclusive), its representation in base b reads the same forwards and backwards. In simpler terms, you need to check all number systems from base 2 up to base n - 2 and ensure that n looks the same when reversed in each of these bases.

The input is a single integer n constrained by 4 <= n <= 10^5. The output is a boolean: true if n is strictly palindromic, false otherwise. Important constraints and insights include that n will always be at least 4, which avoids trivial cases like 1 or 2, and the number of bases to check grows linearly with n. A naive solution might attempt to convert n into every base from 2 to n - 2 and check for palindromes, which can be computationally expensive for large n.

An edge case to be aware of is the smallest valid n = 4. Here, only base 2 needs checking, and 4 in base 2 is "100", which is not palindromic. Another subtle point is that very large numbers near the upper limit 10^5 might require careful handling if using a brute-force conversion for all bases.

Interestingly, there is a mathematical insight that makes this problem trivial: no number n >= 4 can be strictly palindromic. In base n-2, n is always represented as "12", which is never a palindrome. This means that for the given constraints, the answer is always false.

Approaches

The brute-force approach would iterate over all bases b from 2 to n-2, convert n to that base as a string, and check if it is a palindrome. While this works for small n, its time complexity grows quadratically with n because converting a number to base b takes O(log_b n) time, and there are O(n) bases to check. This is inefficient for n near 10^5.

The optimal approach relies on the mathematical property mentioned above. For any n >= 4, converting n to base n-2 yields "12", which is not palindromic. Therefore, without doing any iteration or conversion, we can directly return false.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n log n)	O(log n)	Convert n to every base from 2 to n-2 and check palindrome
Optimal	O(1)	O(1)	Use mathematical property: n >= 4 is never strictly palindromic

Algorithm Walkthrough

1. Receive the input integer n.
2. Check if n is at least 4. Given constraints guarantee this, so no extra check is needed.
3. Return false directly because no number greater than or equal to 4 can satisfy strict palindromicity.

Why it works: In base n-2, the representation of n is always "12", which is not a palindrome. Since strictly palindromic requires the number to be palindromic in all bases from 2 to n-2, the condition fails for n >= 4. This mathematical proof eliminates the need for iteration or conversion.

Python Solution

class Solution:
    def isStrictlyPalindromic(self, n: int) -> bool:
        return False

Explanation: The solution leverages the mathematical insight. The function takes n as input and directly returns False because no number n >= 4 can be strictly palindromic. This avoids unnecessary computations.

Go Solution

func isStrictlyPalindromic(n int) bool {
    return false
}

Explanation: The Go implementation mirrors the Python solution. There is no need for loops or conversions because the property holds for all valid n.

Worked Examples

Example 1: n = 9

Normally, we would check bases 2 through 7:

Base	Representation	Palindrome?
2	1001	Yes
3	100	No

Since base 3 is not palindromic, 9 is not strictly palindromic. The optimal solution immediately returns false.

Example 2: n = 4

Only base 2 is considered:

Base	Representation	Palindrome?
2	100	No

Again, 4 is not strictly palindromic. The optimal solution immediately returns false.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	No computation or iteration is required; the function returns immediately
Space	O(1)	No additional memory is allocated

Even though a brute-force approach would be much slower, the mathematical property allows constant-time evaluation.

Test Cases

# Provided examples
assert Solution().isStrictlyPalindromic(9) == False  # base 3 check fails
assert Solution().isStrictlyPalindromic(4) == False  # base 2 check fails

# Boundary values
assert Solution().isStrictlyPalindromic(5) == False  # smallest n > 4
assert Solution().isStrictlyPalindromic(105) == False  # upper limit

# Random larger number
assert Solution().isStrictlyPalindromic(10000) == False  # still always false

Test	Why
n = 9	Checks base 3 fails
n = 4	Checks smallest valid input fails
n = 5	Confirms next smallest input fails
n = 105	Confirms large input fails
n = 10000	Validates function handles large inputs efficiently

Edge Cases

The first edge case is n = 4. This is the smallest allowed value and ensures that the solution correctly identifies that strict palindromicity fails for minimal inputs. The function returns false without error.

The second edge case is n close to the upper bound, such as n = 10^5. A brute-force approach would take a long time and could lead to timeouts. The optimal solution bypasses this completely.

The third edge case is odd or prime numbers. Regardless of number type, the property that n in base n-2 is "12" guarantees the result is always false. This avoids subtle bugs that could arise if someone attempted to check palindromicity manually for all bases.