Leetcode 2396: Strictly Palindromic Number
An integer
n is considered strictly palindromic if, for every base b between 2 and n - 2 (inclusive), the representation of n in base b is a palindrome. A number is palindromic in a base if the string representation of that number in that base reads the same forward and backward. Your task is to determine if the given integer n is strictly palindromic or not.Problem
Approach
Steps
Complexity
Input: The input consists of a single integer `n` (4 <= n <= 10^5), where `n` is the number you need to check for strict palindromicity across various bases.
Example: n = 5
Constraints:
• 4 <= n <= 10^5
Output: Return `true` if `n` is strictly palindromic, otherwise return `false`.
Example: Output: false
Constraints:
Goal: The goal is to check if for all bases between 2 and `n - 2`, the base `b` representation of `n` is palindromic. If for any base it is not palindromic, return false.
Steps:
• 1. Iterate through all bases from 2 to `n - 2`.
• 2. For each base, convert `n` to its string representation in that base.
• 3. Check if the string representation is a palindrome.
• 4. If a non-palindromic representation is found, return false.
• 5. If all representations are palindromic, return true.
Goal: The solution must work within the time limits for large values of `n`, and correctly handle the representation of `n` in bases ranging from 2 to `n-2`.
Steps:
• Ensure the solution can handle values of `n` up to 10^5 efficiently.
Assumptions:
• The number `n` is always greater than or equal to 4.
• Input: n = 6
• Explanation: In base 2: 6 = 110 (not palindromic), base 3: 6 = 20 (not palindromic), base 4: 6 = 12 (not palindromic), hence 6 is not strictly palindromic.
• Input: n = 7
• Explanation: In base 2: 7 = 111 (palindromic), base 3: 7 = 21 (not palindromic), hence 7 is not strictly palindromic.
Approach: The approach is to iterate over all bases between 2 and `n - 2`, converting the number `n` to its base `b` representation and checking if the representation is palindromic. If any base has a non-palindromic representation, the function will return false.
Observations:
• This problem involves converting numbers to various bases and checking for palindromic properties.
• The challenge lies in efficiently checking palindromes for all bases up to `n-2`.
Steps:
• 1. Initialize a loop that iterates through each base from 2 to `n - 2`.
• 2. For each base, convert `n` to its representation in that base.
• 3. Check if the string representation of `n` in that base is a palindrome.
• 4. If any non-palindromic representation is found, return false.
• 5. If all representations are palindromic, return true.
Empty Inputs:
• Since `n` is always >= 4, there are no empty inputs.
Large Inputs:
• Handle cases where `n` is near the upper limit of 10^5.
Special Values:
• Consider how values near the lower bound (like `n = 4`) behave.
Constraints:
• Make sure that the solution is efficient enough to handle the largest inputs within the time limit.
bool isStrictlyPalindromic(int n) {
return false;
}
1 : Function Declaration
bool isStrictlyPalindromic(int n) {
The function `isStrictlyPalindromic` is declared. It takes an integer `n` as input and is expected to return a boolean value indicating whether `n` is strictly palindromic in a given base.
2 : Return Statement
return false;
The function currently returns `false`, meaning that no number is considered strictly palindromic in this implementation. The logic for palindromic checking in various bases needs to be implemented.
Best Case: O(n), where n is the length of the input number.
Average Case: O(n), where n is the length of the input number.
Worst Case: O(n^2), where n is the length of the input number due to base conversion and palindrome checking.
Description: The worst-case time complexity is O(n^2) because for each base we convert the number and check if it's palindromic.
Best Case: O(1), if using a constant amount of extra space.
Worst Case: O(n), where n is the length of the input number due to space used in storing the string representations.
Description: The space complexity is O(n) in the worst case, where n is the number of digits in the base `b` representation of `n`.
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