LeetCode 866: Prime Palindrome

Explanation:

To find the smallest prime palindrome greater than or equal to a given number n, we can iterate through numbers starting from n and check if each number is a palindrome and a prime number. We can implement functions to check for palindromes and prime numbers, and then use these functions to find the desired number.

Algorithm:

1. Start iterating from n and check each number if it is a palindrome and a prime number.
2. Check if a number is a palindrome by converting it to a string and comparing the reverse string with the original string.
3. Check if a number is a prime number by iterating from 2 to the square root of the number and checking if it has any divisors.
4. Return the first number that satisfies both conditions.

Time Complexity:

The time complexity of this approach is O(sqrt(n) * log(n)), where n is the input number.

Space Complexity:

The space complexity of this approach is O(1).

: :

class Solution {
    public int primePalindrome(int n) {
        while (true) {
            if (isPalindrome(n) && isPrime(n)) {
                return n;
            }
            n++;
            if (10000000 < n && n < 100000000) {
                n = 100000000;
            }
        }
    }
    
    private boolean isPalindrome(int num) {
        String str = String.valueOf(num);
        return new StringBuilder(str).reverse().toString().equals(str);
    }
    
    private boolean isPrime(int num) {
        if (num < 2) {
            return false;
        }
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }
}