Prime Palindrome

Prime Palindrome - Problem

Given an integer n, return the smallest prime palindrome greater than or equal to n.

An integer is prime if it has exactly two divisors: 1 and itself. Note that 1 is not a prime number. For example, 2, 3, 5, 7, 11, and 13 are all primes.

An integer is a palindrome if it reads the same from left to right as it does from right to left. For example, 101 and 12321 are palindromes.

The test cases are generated so that the answer always exists and is in the range [2, 2 * 10⁸].

Input & Output

Example 1 — Small Number

$ Input: n = 9

› Output: 11

💡 Note: Starting from 9: 9 is palindrome but not prime (divisible by 3), 10 is not palindrome, 11 is both palindrome and prime (only divisible by 1 and 11)

Example 2 — Already Prime Palindrome

$ Input: n = 2

› Output: 2

💡 Note: 2 itself is both prime and palindrome, so return 2

Example 3 — Larger Input

$ Input: n = 13

› Output: 101

💡 Note: 13 is prime but not palindrome. Next palindromes are 22, 33, 44, etc. but they're not prime. 101 is the first prime palindrome ≥ 13

Constraints

1 ≤ n ≤ 2 × 10⁸
The answer always exists and is in the range [2, 2 × 10⁸]

Visualization

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Asked in

G Google 15 F Facebook 10 a Amazon 8

The key insight is that even-length palindromes (except 11) are divisible by 11, so we focus on odd-length palindromes. The optimal approach generates palindrome candidates systematically rather than checking every number. Best approach: Generate palindromes by mirroring first half, then test primality. Time: O(√n × log n), Space: O(log n)

Common Approaches

✓ Brute Force Check

⏱️ Time: O(n × √n) Space: O(1)

Starting from n, check each number to see if it's both a palindrome and prime. Return the first number that satisfies both conditions.

Generate Palindromes

⏱️ Time: O(√n × log n) Space: O(log n)

Instead of checking every number, generate palindromes by constructing them from their first half, then test only these candidates for primality. Skip even-length palindromes except 11.

Brute Force Check — Algorithm Steps

Start from n
For each number, check if it's a palindrome
If palindrome, check if it's prime
Return first number that is both

Visualization

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Step-by-Step Walkthrough

Start

Begin checking from n = 9

Check

Test each number for palindrome and prime

Found

Return first number that satisfies both conditions

Code -

solution.c — C

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

int isPalindrome(int num) {
    char s[20];
    sprintf(s, "%d", num);
    int len = strlen(s);
    for (int i = 0; i < len / 2; i++) {
        if (s[i] != s[len - 1 - i]) {
            return 0;
        }
    }
    return 1;
}

int isPrime(int num) {
    if (num < 2) return 0;
    if (num == 2) return 1;
    if (num % 2 == 0) return 0;
    for (int i = 3; i * i <= num; i += 2) {
        if (num % i == 0) return 0;
    }
    return 1;
}

int solution(int n) {
    int current = n;
    while (1) {
        if (isPalindrome(current) && isPrime(current)) {
            return current;
        }
        current++;
    }
}

int main() {
    int n;
    scanf("%d", &n);
    printf("%d\n", solution(n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × √n)

In worst case, check up to n numbers, each prime check takes O(√n) time

✓ Linear Growth

Space Complexity

O(1)

Only use constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Check — Algorithm Steps

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Code -

Time & Space Complexity

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