866. 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 * 10^8].

  Example 1:

Input: n = 6
Output: 7

Example 2:

Input: n = 8
Output: 11

Example 3:

Input: n = 13
Output: 101

  Constraints:

• 1 <= n <= 10^8

Solution (Java)

class Solution {
    private boolean isPrime(int n) {
        if (n % 2 == 0) {
            return n == 2;
        }
        for (int i = 3, s = (int) Math.sqrt(n); i <= s; i += 2) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    private int next(int num) {
        char[] s = String.valueOf(num + 1).toCharArray();
        for (int i = 0, n = s.length; i < (n >> 1); i++) {
            while (s[i] != s[n - 1 - i]) {
                increment(s, n - 1 - i);
            }
        }
        return Integer.parseInt(new String(s));
    }

    private void increment(char[] s, int i) {
        while (s[i] == '9') {
            s[i--] = '0';
        }
        s[i]++;
    }

    public int primePalindrome(int n) {
        if (n <= 2) {
            return 2;
        }
        int p = next(n - 1);
        while (!isPrime(p)) {
            if (10_000_000 <= p && p < 100_000_000) {
                p = 100_000_000;
            }
            p = next(p);
        }
        return p;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).