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1175. Prime Arrangements

Description

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

Solutions

Solution 1: Mathematics

First, count the number of prime numbers within the range $[1,n]$, which we denote as $cnt$. Then, calculate the product of the factorial of $cnt$ and $n-cnt$ to get the answer, remember to perform the modulo operation.

Here, we use the “Sieve of Eratosthenes” to count prime numbers.

If $x$ is a prime number, then multiples of $x$ greater than $x$, such as $2x$, $3x$, … are definitely not prime numbers, so we can start from here.

Let $primes[i]$ indicate whether the number $i$ is a prime number. If it is a prime number, it is $true$, otherwise it is $false$.

We sequentially traverse each number $i$ in the range $[2,n]$. If this number is a prime number, the number of prime numbers increases by $1$, and then all its multiples $j$ are marked as composite numbers (except for the prime number itself), that is, $primes[j]=false$. In this way, at the end of the run, we can know the number of prime numbers.

The time complexity is $O(n\times \log \log n)$.

• Java
• C++
• Python
• Go

• class Solution {
      private static final int MOD = (int) 1e9 + 7;
  
      public int numPrimeArrangements(int n) {
          int cnt = count(n);
          long ans = f(cnt) * f(n - cnt);
          return (int) (ans % MOD);
      }
  
      private long f(int n) {
          long ans = 1;
          for (int i = 2; i <= n; ++i) {
              ans = (ans * i) % MOD;
          }
          return ans;
      }
  
      private int count(int n) {
          int cnt = 0;
          boolean[] primes = new boolean[n + 1];
          Arrays.fill(primes, true);
          for (int i = 2; i <= n; ++i) {
              if (primes[i]) {
                  ++cnt;
                  for (int j = i + i; j <= n; j += i) {
                      primes[j] = false;
                  }
              }
          }
          return cnt;
      }
  }
  

• using ll = long long;
  const int MOD = 1e9 + 7;
  
  class Solution {
  public:
      int numPrimeArrangements(int n) {
          int cnt = count(n);
          ll ans = f(cnt) * f(n - cnt);
          return (int) (ans % MOD);
      }
  
      ll f(int n) {
          ll ans = 1;
          for (int i = 2; i <= n; ++i) ans = (ans * i) % MOD;
          return ans;
      }
  
      int count(int n) {
          vector<bool> primes(n + 1, true);
          int cnt = 0;
          for (int i = 2; i <= n; ++i) {
              if (primes[i]) {
                  ++cnt;
                  for (int j = i + i; j <= n; j += i) primes[j] = false;
              }
          }
          return cnt;
      }
  };
  

• class Solution:
      def numPrimeArrangements(self, n: int) -> int:
          def count(n):
              cnt = 0
              primes = [True] * (n + 1)
              for i in range(2, n + 1):
                  if primes[i]:
                      cnt += 1
                      for j in range(i + i, n + 1, i):
                          primes[j] = False
              return cnt
  
          cnt = count(n)
          ans = factorial(cnt) * factorial(n - cnt)
          return ans % (10**9 + 7)
  
  

• func numPrimeArrangements(n int) int {
  	count := func(n int) int {
  		cnt := 0
  		primes := make([]bool, n+1)
  		for i := range primes {
  			primes[i] = true
  		}
  		for i := 2; i <= n; i++ {
  			if primes[i] {
  				cnt++
  				for j := i + i; j <= n; j += i {
  					primes[j] = false
  				}
  			}
  		}
  		return cnt
  	}
  
  	mod := int(1e9) + 7
  	f := func(n int) int {
  		ans := 1
  		for i := 2; i <= n; i++ {
  			ans = (ans * i) % mod
  		}
  		return ans
  	}
  
  	cnt := count(n)
  	ans := f(cnt) * f(n-cnt)
  	return ans % mod
  }
  

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