g0801_0900.s0866_prime_palindrome.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g0801_0900.s0866_prime_palindrome;

// #Medium #Math #2022_03_27_Time_2_ms_(84.68%)_Space_41_MB_(71.89%)

/**
 * 866 - Prime Palindrome\.
 *
 * Medium
 *
 * 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].
 *
 * **Example 1:**
 *
 * **Input:** n = 6
 *
 * **Output:** 7
 *
 * **Example 2:**
 *
 * **Input:** n = 8
 *
 * **Output:** 11
 *
 * **Example 3:**
 *
 * **Input:** n = 13
 *
 * **Output:** 101
 *
 * **Constraints:**
 *
 * *   1 <= n <= 10⁸
**/
public 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;
    }
}