Formatted question description: https://leetcode.ca/all/866.html

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/

// 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/

// 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;
        }
    }
};

Java

class Solution {
    public int primePalindrome(int N) {
        int palindromeNum = nextPalindrome(N);
        while (!isPrime(palindromeNum))
            palindromeNum = nextPalindrome(palindromeNum + 1);
        return palindromeNum;
    }

    public int nextPalindrome(int num) {
        int length = String.valueOf(num).length();
        if (num + 1 == (int) Math.pow(10, length))
            return num + 2;
        int divisor = (int) Math.pow(10, length / 2);
        int headerNum = num / divisor;
        boolean even = length % 2 == 0;
        int palindromeNum = generatePalindrome(headerNum, even);
        if (palindromeNum < num)
            palindromeNum = generatePalindrome(headerNum + 1, even);
        return palindromeNum;
    }

    public int generatePalindrome(int num, boolean even) {
        StringBuffer sb1 = new StringBuffer(String.valueOf(num));
        StringBuffer sb2 = new StringBuffer(String.valueOf(num));
        sb2.reverse();
        if (!even)
            sb2.deleteCharAt(0);
        String palindromeStr = new StringBuffer(sb1.toString()).append(sb2.toString()).toString();
        return Integer.parseInt(palindromeStr);
    }

    public boolean isPrime(int num) {
        if (num == 1)
            return false;
        if (num == 2 || num == 3)
            return true;
        if (num % 2 == 0 || num % 3 == 0)
            return false;
        int sqrt = (int) Math.sqrt(num);
        for (int i = 5; i <= sqrt; i += 2) {
            if (num % i == 0)
                return false;
        }
        return true;
    }
}

// OJ: https://leetcode.com/problems/prime-palindrome/
// Time: O(N)
// Space: O(logN)
// Ref: https://leetcode.com/problems/prime-palindrome/solution/
class Solution {
    int getPalindrome(int half, bool odd) {
        string s = to_string(half);
        for (int i = s.size() - 1 - odd; i >= 0; --i) s += s[i];
        return stoi(s);
    }
    bool isPrime(int n) {
        if (n < 2) return false;
        for (int d = 2; d * d <= n; ++d) {
            if (n % d == 0) return false;
        }
        return true;
    }
public:
    int primePalindrome(int n) {
        for (int len = 1; len <= 5; ++len) {
            for (int root = pow(10, len - 1); root < pow(10, len); ++root) { // Enumerate odd-length palindromes
                int pal = getPalindrome(root, true);
                if (pal >= n && isPrime(pal)) return pal;
            }
            for (int root = pow(10, len - 1); root < pow(10, len); ++root) { // Enumerate even-length palindromes
                int pal = getPalindrome(root, false);
                if (pal >= n && isPrime(pal)) return pal;
            }
        }
        return -1;
    }
};

```
print("Todo!")
```

866 - Prime Palindrome

866. Prime Palindrome (Medium)

Solution 1.

Solution 2. Brute Force with Mathematical Shortcut

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