866. Prime Palindrome

866. Prime Palindrome (Medium)

Find the smallest prime palindrome greater than or equal to N.

Recall that a number is prime if it's only divisors are 1 and itself, and it is greater than 1. 

For example, 2,3,5,7,11 and 13 are primes.

Recall that a number is a palindrome if it reads the same from left to right as it does from right to left. 

For example, 12321 is a palindrome.

 

Example 1:

Input: 6
Output: 7

Example 2:

Input: 8
Output: 11

Example 3:

Input: 13
Output: 101

 

Note:

• 1 <= N <= 10^8
• The answer is guaranteed to exist and be less than 2 * 10^8.

Related Topics:
Math

Solution 1.

// OJ: https://leetcode.com/problems/prime-palindrome/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(logN)
// Ref: https://leetcode.com/problems/prime-palindrome/solution/
class Solution {
    bool isPrime(int x) {
        if (x < 2) return false;
        for (int d = 2, R = sqrt(x); d <= R; ++d) {
            if (x % d == 0) return false;
        }
        return true;
    }
public:
    int primePalindrome(int N) {
        for (int L = 1; L <= 5; ++L) {
            for (int root = pow(10, L - 1); root < pow(10, L); ++root) { // check for odd-length palindromes
                auto s = to_string(root);
                for (int k = L - 2; k >= 0; --k) s += s[k];
                int x = stoi(s);
                if (x >= N && isPrime(x)) return x;
            }
            for (int root = pow(10, L - 1); root < pow(10, L); ++root) {
                auto s = to_string(root);
                for (int k = L - 1; k >= 0; --k) s += s[k];
                int x = stoi(s);
                if (x >= N && isPrime(x)) return x;
            }
        }
        return -1;
    }
};

Solution 2. Brute Force with Mathematical Shortcut

// OJ: https://leetcode.com/problems/prime-palindrome/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/prime-palindrome/solution/
class Solution {
    bool isPrime(int x) {
        if (x < 2) return false;
        for (int d = 2, R = sqrt(x); d <= R; ++d) {
            if (x % d == 0) return false;
        }
        return true;
    }
    int reverse(int N) {
        int ans = 0;
        while (N > 0) {
            ans = 10 * ans + (N % 10);
            N /= 10;
        }
        return ans;
    }
public:
    int primePalindrome(int N) {
        while (true) {
            if (N == reverse(N) && isPrime(N)) return N;
            ++N;
            if (1e7 < N && N < 1e8) N = 1e8;
        }
    }
};