Strictly Palindromic Number

Strictly Palindromic Number - Problem

An integer n is strictly palindromic if, for every base b between 2 and n - 2 (inclusive), the string representation of the integer n in base b is palindromic.

Given an integer n, return true if n is strictly palindromic and false otherwise.

A string is palindromic if it reads the same forward and backward.

Input & Output

Example 1 — Small Number

$ Input: n = 9

› Output: false

💡 Note: In base 2: 9 = 1001 (palindromic), but in base 3: 9 = 100 (not palindromic), so return false

Example 2 — Very Small Number

$ Input: n = 4

› Output: false

💡 Note: Need to check bases [2]. In base 2: 4 = 100 (not palindromic), so return false

Example 3 — Edge Case

$ Input: n = 2

› Output: false

💡 Note: No bases to check (range [2, 0] is empty), but by definition return false

Constraints

1 ≤ n ≤ 2 × 10⁹

Visualization

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Asked in

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This is essentially a trick question with a mathematical twist. The key insight is that no integer can be strictly palindromic due to the constraint requirements. The optimal approach simply returns false in O(1) time and O(1) space, making it one of the most efficient problems possible.

Common Approaches

✓ Brute Force - Check All Bases

⏱️ Time: O(n² log n) Space: O(log n)

For each base b from 2 to n-2, convert n to base b representation and check if the resulting string reads the same forwards and backwards. If any base produces a non-palindromic representation, return false.

Mathematical Insight - Direct False

⏱️ Time: O(1) Space: O(1)

Based on mathematical analysis, for any n > 4, there exists at least one base in the range [2, n-2] where n's representation is not palindromic. This means we can immediately return false without any computation.

Brute Force - Check All Bases — Algorithm Steps

Step 1: Iterate through all bases from 2 to n-2
Step 2: Convert n to each base b representation
Step 3: Check if the representation is palindromic using two pointers
Step 4: Return false if any base fails, true if all pass

Visualization

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Step-by-Step Walkthrough

Convert to Bases

Convert n=9 to base 2,3,4,5,6,7

Check Palindromes

Verify each representation reads same forwards/backwards

Result

Return false if any base fails palindrome test

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

void toBase(int num, int base, int* digits, int* len) {
    if (num == 0) {
        digits[0] = 0;
        *len = 1;
        return;
    }
    
    *len = 0;
    int temp = num;
    while (temp > 0) {
        digits[(*len)++] = temp % base;
        temp /= base;
    }
    
    // Reverse the array
    for (int i = 0; i < *len / 2; i++) {
        int swap = digits[i];
        digits[i] = digits[*len - 1 - i];
        digits[*len - 1 - i] = swap;
    }
}

bool isPalindrome(int* arr, int len) {
    int left = 0, right = len - 1;
    while (left < right) {
        if (arr[left] != arr[right]) return false;
        left++;
        right--;
    }
    return true;
}

bool solution(int n) {
    if (n <= 3) return false;
    
    int digits[32];
    int len;
    
    for (int base = 2; base < n - 1; base++) {
        toBase(n, base, digits, &len);
        if (!isPalindrome(digits, len)) {
            return false;
        }
    }
    
    return true;
}

int main() {
    int n;
    scanf("%d", &n);
    bool result = solution(n);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² log n)

For each of n bases, we do O(log n) work to convert and O(log n) to check palindrome

✓ Linear Growth

Space Complexity

O(log n)

Storage for the base representation digits

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Bases — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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