Formatted question description: https://leetcode.ca/all/1175.html

# 1175. Prime Arrangements (Easy)

Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.)

(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)

Since the answer may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: n = 5
Output: 12
Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.


Example 2:

Input: n = 100
Output: 682289015


Constraints:

• 1 <= n <= 100

Related Topics:
Math

## Solution 1. Sieve of Eratosthenes

// OJ: https://leetcode.com/problems/prime-arrangements/
// Time: O(NloglogN)
// Space: O(N)
class Solution {
int countPrimes(int n) {
if (n <= 1) return 0;
vector<int> prime(n + 1, true);
int ans = 0, bound = sqrt(n);
for (int i = 2; i <= n; ++i) {
if (!prime[i]) continue;
++ans;
if (i > bound) continue;
for (int j = i * i; j <= n; j += i) prime[j] = false;
}
return ans;
}
public:
int numPrimeArrangements(int n) {
int cnt = countPrimes(n), ans = 1, mod = 1e9 + 7;
for (int i = 1; i <= cnt; ++i) ans = ((long)ans * i) % mod;
for (int i = 1; i <= n - cnt; ++i) ans = ((long)ans * i) % mod;
return ans;
}
};


Java

• class Solution {
public int numPrimeArrangements(int n) {
final int MODULO = 1000000007;
int primeCount = 0;
for (int i = 1; i <= n; i++) {
if (isPrime(i))
primeCount++;
}
int notPrimeCount = n - primeCount;
long arrangements = 1L;
for (long i = 1L; i <= primeCount; i++)
arrangements = (arrangements * i) % MODULO;
for (long i = 1L; i <= notPrimeCount; i++)
arrangements = (arrangements * i) % MODULO;
return (int) arrangements;
}

public boolean isPrime(int num) {
if (num == 1)
return false;
if (num == 2 || num == 3 || num == 5 || num == 7)
return true;
if (num % 2 == 0 || num % 3 == 0 || num % 5 == 0 || num % 7 == 0)
return false;
int sqrt = (int) Math.sqrt(num);
for (int i = 11; i <= sqrt; i += 2) {
if (num % i == 0)
return false;
}
return true;
}
}

• // OJ: https://leetcode.com/problems/prime-arrangements/
// Time: O(NloglogN)
// Space: O(N)
class Solution {
int countPrimes(int n) {
if (n <= 1) return 0;
vector<int> prime(n + 1, true);
int ans = 0, bound = sqrt(n);
for (int i = 2; i <= n; ++i) {
if (!prime[i]) continue;
++ans;
if (i > bound) continue;
for (int j = i * i; j <= n; j += i) prime[j] = false;
}
return ans;
}
public:
int numPrimeArrangements(int n) {
int cnt = countPrimes(n), ans = 1, mod = 1e9 + 7;
for (int i = 1; i <= cnt; ++i) ans = ((long)ans * i) % mod;
for (int i = 1; i <= n - cnt; ++i) ans = ((long)ans * i) % mod;
return ans;
}
};

• # 1175. Prime Arrangements
# https://leetcode.com/problems/prime-arrangements

class Solution:
def numPrimeArrangements(self, n: int) -> int:
primes = [True] * (n + 1)

for prime in range(2, int(math.sqrt(n)) + 1):
if primes[prime]:
for composite in range(prime * prime, n + 1, prime):
primes[composite] = False

cnt = sum(primes[2:])
​
return math.factorial(cnt) * math.factorial(n - cnt) % (10**9 + 7)