Prime Permutation Power: Counting Prime Arrangements Using C++
Intuition\nTo solve the problem of counting prime arrangements, we first need to identify how many prime numbers exist up to a given number n. The problem revol
# Intuition\nTo solve the problem of counting prime arrangements, we first need to identify how many prime numbers exist up to a given number `n`. The problem revolves around arranging primes and non-primes in factorial ways, so understanding the distinction between primes and non-primes is crucial.\n\n# Approach\n1. **Prime Identification**: Create a helper function `prime(n)` that checks if a number `n` is prime. The function returns `1` if the number is prime and `0` otherwise.\n2. **Counting Primes**: Iterate through all numbers from `1` to `n` and count how many of them are prime.\n3. **Calculate Permutations**: \n   - Calculate the factorial of the prime count modulo `1000000007`.\n   - Similarly, calculate the factorial of the non-prime count modulo `1000000007`.\n   - The result is the product of these two factorials.\n\n# Complexity\n- **Time complexity**: $$O(n \\sqrt{n})$$ \u2014 Checking for primality up to $$n$$ takes $$O(\\sqrt{n})$$ time for each number, and there are $$n$$ numbers to check.\n- **Space complexity**: $$O(1)$$ \u2014 Only a constant amount of extra space is used, excluding the input and output.\n\n# Code\n```cpp\nclass Solution {\npublic:\n    int prime(int n) {\n        if (n == 1) {\n            return 0; \n        }\n        for (int i = 2; i * i <= n; ++i) {\n            if (n % i == 0) {\n                return 0; \n            }\n        }\n        return 1; \n    }\n\n    int numPrimeArrangements(int n) {\n        int count = 0; \n        for (int i = 1; i <= n; ++i) {\n            if (prime(i) == 1) {\n                count++; \n            }\n        }\n        long long int perm = 1; \n        for (int i = 1; i <= count; ++i) {\n            perm = (perm * i) % 1000000007; \n        }\n        for (int i = 1; i <= n - count; ++i) {\n            perm = (perm * i) % 1000000007; \n        }\n        return perm; \n    }\n};\n```\n
['C++']