LeetCode 2396: Strictly Palindromic Number Solution

Master LeetCode problem 2396 (Strictly Palindromic Number), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2396. Strictly Palindromic Number

Medium

Math Two Pointers Brainteaser

Problem Explanation

Explanation:

To solve this problem, we need to check if a given number n is strictly palindromic. A number is strictly palindromic if, for every base b between 2 and n - 2 inclusive, the string representation of n in base b is palindromic.

We can iterate over the bases from 2 to n - 2 and convert the number n to each base to check if it is a palindrome in that base. If we find a base where n is not a palindrome, we can immediately return false. If we successfully iterate over all bases without finding any base where n is not a palindrome, we return true.

Algorithm:

Iterate over bases from 2 to n - 2.
Convert number n to the current base.
Check if the converted number is a palindrome.
If any base conversion is not a palindrome, return false. Otherwise, return true.

Time Complexity: The time complexity of this algorithm is O(log2(n) * n) since we iterate over each base from 2 to n - 2 and convert the number n to each base.

Space Complexity: The space complexity is O(log2(n)) to store the string representation of the number in different bases.

Solution Code

class Solution {
    public boolean isStrictlyPalindromic(int n) {
        for (int base = 2; base <= n - 2; base++) {
            String numInBase = Integer.toString(n, base);
            if (!isPalindrome(numInBase)) {
                return false;
            }
        }
        return true;
    }
    
    private boolean isPalindrome(String s) {
        int left = 0, right = s.length() - 1;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

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