Simple C# Solution (With Explanation)
Approach\n\nThe code finds the smallest prime palindrome number greater than or equal to a given integer (n . Here\'s how it works step by step:\n\n1. Edge Case
# Approach\n\nThe code finds the smallest prime palindrome number greater than or equal to a given integer (n . Here\'s how it works step by step:\n\n1. **Edge Case for Small Values**:  \n   If (n <= 2), the function directly returns `2` since it is the smallest prime palindrome.\n\n2. **Iterative Search**:  \n   Starting from (n), the function iterates to find a number that satisfies two conditions:\n   - It is a **palindrome** (reads the same forward and backward).\n   - It is **prime** (divisible only by 1 and itself).\n\n3. **Optimization with Odd Numbers**:  \n   - The function skips even numbers (except for (2)) since even numbers (greater than (2)) cannot be prime.\n\n4. **Palindrome Check First**:  \n   - The function first checks if a number is a palindrome using the `IsPalindrome` method. This avoids unnecessary primality checks for non-palindromic numbers.\n\n5. **Prime Check Next**:  \n   - If the number is a palindrome, the `IsPrime` function verifies its primality.\n\n6. **Result**:  \n   - Once a number satisfies both conditions, it is returned as the result.\n\n\n# Complexity\n- **Time complexity: O(k * sqrt(n))**\n    - **IsPalindrome Function**:\n        - The `IsPalindrome` function has a time complexity of (O(d)), where (d) is the number of digits in the number.\n        - For a number (n), (d) is proportional to (log10(n)). This is generally very small compared to the cost of the `IsPrime` function and does not dominate the complexity.\n\n    - **IsPrime Function**:\n        - The `IsPrime` function checks divisors up to (sqrt(n)), so its complexity is ( O(sqrt(n))) for each call.\n\n    - **PrimePalindrome Function**:\n        - The `PrimePalindrome` function iterates through numbers, skipping even numbers and checking each number for both palindromic property and primality.\n        - If the density of prime palindromes is very low (which it generally is), the loop could potentially iterate through many numbers, performing `IsPalindrome` and `IsPrime` checks repeatedly.\n\n    - **Why \\( O(n \\cdot \\sqrt{n}) \\) is an Overestimate**\n\n        - The **key factor** here is how many numbers the loop iterates through before finding a prime palindrome.\n        - The loop doesn\'t iterate through all \\( n \\) numbers because:\n        - Non-palindromic numbers are skipped as soon as `IsPalindrome` fails.\n        - Many composite numbers are skipped because `IsPrime` will terminate early.\n  \n    - Thus, the iteration count is proportional to the sparsity of **prime palindromes**, which is far lower than the sparsity of primes themselves.\n\n    - **Refined Time Complexity**\n        - The actual complexity is closer to:\n            - **Best Case**: (O(sqrt(n)))) (if (n) itself is a prime palindrome).\n            - **Average Case**: Proportional to the density of prime palindromes. Empirically, this is lower than (O(n * sqrt(n)))).\n            - **Worst Case**: (O(k * sqrt(n))), where (k) is the number of numbers checked before finding the result. (k) depends on the sparsity of prime palindromes.\n\n        - While `O(n * sqrt(n)))` is a theoretical upper bound, the actual practical complexity is lower due to the filtering done by `IsPalindrome`.\n\n- **Space Complexity: O(1)**\n    - The algorithm uses a few variables and performs checks without storing large structures.\n\n# Code\n```csharp []\npublic class Solution {\n    public int PrimePalindrome(int n) {\n        int num = n;\n\n        if(n <= 2)\n        {\n            return 2;\n        }\n        string s;\n        while(true)\n        {\n            // This condition is added for optimization.\n            if(num % 2 == 0)\n            {\n                num++;\n                continue;\n            }\n            // Important is to check IsPalindrome first & then IsPrime.\n            if(IsPalindrome(num) && IsPrime(num))\n            {\n                return num;\n            }\n            num++;\n        }\n        return num;\n    }\n    private bool IsPrime(int n)\n    {\n        int sqrt = (int)Math.Sqrt(n);\n        for(int i=2; i<=sqrt; i++)\n        {\n            if(n % i == 0)\n            {\n                return false;\n            }\n        }\n        return true;\n    }\n\n    private bool IsPalindrome(int n)\n    {\n        string s = n.ToString();\n        int left=0, right = s.Length-1;\n        while(left < right)\n        {\n            if(s[left] != s[right])\n            {\n                return false;\n            }\n            left++;\n            right--;\n        }\n        return true;\n    }\n}\n```
['Math', 'Number Theory', 'C#']