Best approach for beginners using Java
IntuitionTo solve this problem, we need to find the smallest prime number that is also a palindrome and greater than or equal to n. A prime number is one that h
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve this problem, we need to find the smallest prime number that is also a palindrome and greater than or equal to n. A prime number is one that has no divisors other than 1 and itself, while a palindrome is a number that reads the same forwards and backwards.\n\n- Starting from n and checking if the current number is both a prime and a palindrome.\n- If it\'s not, we increment the number and check again.\n- A quick optimization involves skipping numbers with even lengths (except for 2) because they cannot be palindromic primes (except for the trivial case of 2), and we can directly jump to the next potential candidate by changing the length of the number.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Start with n: Begin from the given number n and check each number to see if it\'s a palindrome and a prime.\n- Check if it\'s prime: Implement a helper function isPrime() that returns true if the number is prime.\n- Check if it\'s a palindrome: Implement another helper function isPalindrome() that returns true if the number is a palindrome.\n- Skip unnecessary checks: If the length of n is even and greater than 11, we know that palindromes with even digits can\'t be prime (except for 2), so we jump to the next possible odd-length palindrome.\n- Repeat until found: Continue this process until we find a number that satisfies both conditions.\n\n\n# Complexity \n- Time complexity: O(k*\u221An)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n    public int primePalindrome(int n) {\n        if (n <= 2) {\n            return 2;\n        }\n\n        while (true) {\n            if (isPrime(n) && isPalindrome(n)) {\n                return n;\n            }\n            n++;\n            if (n > 11 && String.valueOf(n).length() % 2 == 0) {\n                n = (int) Math.pow(10, String.valueOf(n).length());\n            }\n        }\n    }\n\n    private boolean isPalindrome(int n) {\n        int num = n;\n        int rev = 0;\n\n        while (n > 0) {\n            int rem = n % 10;\n            rev = rev * 10 + rem;\n            n /= 10;\n        }\n        return rev == num;\n    }\n\n    private boolean isPrime(int n) {\n        if (n <= 1) {\n            return false;\n        }\n        if (n == 2) {\n            return true;\n        }\n        if (n % 2 == 0) {\n            return false;\n        }\n\n        for (int i = 3; i * i <= n; i += 2) {\n            if (n % i == 0) {\n                return false;\n            }\n        }\n        return true;\n    }\n}\n\n```
['Java']