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Count of subsequences which consists exactly K prime numbers

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Given an integer K and an array arr[], the task is to find the number of subsequences from the given array such that each subsequence consists exactly K prime numbers.
Example: 
 

Input: K = 2, arr = [2, 3, 4, 6] 
Output:
Explanation: 
There are 4 subsequences which consists exactly 2 prime numbers {2, 3} {2, 3, 4} {2, 3, 6} {2, 3, 4, 6}
Input: K = 3, arr = [1, 2, 3, 4, 5, 6, 7] 
Output: 32 
Explanation: 
There are 32 subsequences which consists exactly 3 prime numbers. 
 


Approach: To solve the problem mentioned above we have to find the count of prime numbers in the given array
 

  1. Let the count of prime numbers is m.
  2. We can choose any K integers among m prime numbers.
  3. So the possible combination of choosing prime numbers in the subsequence is mCk    and in the subsequence we can add any number of non – prime numbers because there is no restriction on the non prime numbers where count of non prime numbers will be (N – m).
  4. We can choose any subset of the (N – m) for nonprime numbers in our subsequence.
  5. Possibility of choosing all subset of size (N – m) is pow(2, (m – N)).
  6. For generating subsequences, we will multiply prime number possibility with nonprime number possibility. 
     

Sub sequence count = (Count of prime numbers) C (K) * pow(2, count of non-prime numbers)


Below is the implementation of the above approach: 
 

C++

// C++ implementation to find
// the count of subsequences
// which consist exactly K primes
 
#include <bits/stdc++.h>
using namespace std;
 
// Returns factorial of n
int fact(int n)
{
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
 
// Function to return total
// number of combinations
int nCr(int n, int r)
{
    return fact(n)
           / (fact(r)
              * fact(n - r));
}
 
// Function check whether a number
// is prime or not
bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to n-1
    for (int i = 2; i < n; i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function for finding number of subsequences
// which consists exactly K primes
int countSubsequences(int arr[], int n, int k)
{
    int countPrime = 0;
    for (int i = 0; i < n; i++) {
        if (isPrime(arr[i]))
            countPrime++;
    }
    // if number of primes are less than k
    if (countPrime < k)
        return 0;
 
    return nCr(countPrime, k)
           * pow(2, (n - countPrime));
}
 
// Driver code
int main()
{
 
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int K = 3;
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countSubsequences(arr, n, K);
    return 0;
}

                    

Java

// Java implementation to find the
// count of subsequences which
// consist exactly K primes
import java.util.*;
 
class GFG{
     
// Returns factorial of n
static int fact(int n)
{
    int res = 1;
    for(int i = 2; i <= n; i++)
        res = res * i;
         
    return res;
}
 
// Function to return total
// number of combinations
static int nCr(int n, int r)
{
    return fact(n) / (fact(r) *
                      fact(n - r));
}
 
// Function check whether a number
// is prime or not
static boolean isPrime(int n)
{
     
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to n-1
    for(int i = 2; i < n; i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function for finding number of subsequences
// which consists exactly K primes
static int countSubsequences(int arr[],
                             int n, int k)
{
    int countPrime = 0;
    for(int i = 0; i < n; i++)
    {
        if (isPrime(arr[i]))
            countPrime++;
    }
     
    // If number of primes are less than k
    if (countPrime < k)
        return 0;
 
    return nCr(countPrime, k) *
          (int)Math.pow(2, (n - countPrime));
}
 
// Driver code
public static void main(String args[])
{
 
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int K = 3;
    int n = arr.length;
 
    System.out.println(countSubsequences(arr, n, K));
}
}
 
// This code is contributed by ANKITKUMAR34

                    

Python3

# Python3 implementation to find the
# count of subsequences which consist
# exactly K primes
 
# Returns factorial of n
def fact(n):
     
    res = 1;
    for i in range(2, n + 1):
        res = res * i
         
    return res
 
# Function to return total
# number of combinations
def nCr(n, r):
     
    return (fact(n) // (fact(r) *
                        fact(n - r)))
 
# Function check whether a number
# is prime or not
def isPrime(n):
     
    # Corner case
    if (n <= 1):
        return False;
 
    # Check from 2 to n-1
    for i in range(2, n):
        if (n % i == 0):
            return False
 
    return True
 
# Function for finding number of subsequences
# which consists exactly K primes
def countSubsequences(arr, n, k):
     
    countPrime = 0
    for i in range(n):
        if (isPrime(arr[i])):
            countPrime += 1
             
    # If number of primes are less than k
    if (countPrime < k):
        return 0
 
    return (nCr(countPrime, k) *
    pow(2, (n - countPrime)))
 
# Driver code
arr = [ 1, 2, 3, 4, 5, 6, 7 ]
K = 3
n = len(arr)
 
print(countSubsequences(arr, n, K))
 
# This code is contributed by ANKITKUMAR34

                    

C#

// C# implementation to find the
// count of subsequences which
// consist exactly K primes
using System;
 
class GFG{
     
// Returns factorial of n
static int fact(int n)
{
    int res = 1;
    for(int i = 2; i <= n; i++)
        res = res * i;
         
    return res;
}
 
// Function to return total
// number of combinations
static int nCr(int n, int r)
{
    return fact(n) / (fact(r) *
                      fact(n - r));
}
 
// Function check whether a number
// is prime or not
static bool isPrime(int n)
{
     
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to n-1
    for(int i = 2; i < n; i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function for finding number of subsequences
// which consists exactly K primes
static int countSubsequences(int []arr,
                             int n, int k)
{
    int countPrime = 0;
    for(int i = 0; i < n; i++)
    {
        if (isPrime(arr[i]))
            countPrime++;
    }
     
    // If number of primes are less than k
    if (countPrime < k)
        return 0;
 
    return nCr(countPrime, k) *
          (int)Math.Pow(2, (n - countPrime));
}
 
// Driver code
public static void Main(String []args)
{
 
    int []arr = { 1, 2, 3, 4, 5, 6, 7 };
    int K = 3;
    int n = arr.Length;
 
    Console.WriteLine(countSubsequences(arr, n, K));
}
}
 
// This code is contributed by gauravrajput1

                    

Javascript

<script>
 
// Javascript implementation to find
// the count of subsequences
// which consist exactly K primes
 
// Returns factorial of n
function fact(n)
{
    var res = 1;
    for (var i = 2; i <= n; i++)
        res = res * i;
    return res;
}
 
// Function to return total
// number of combinations
function nCr(n, r)
{
    return fact(n)
           / (fact(r)
              * fact(n - r));
}
 
// Function check whether a number
// is prime or not
function isPrime(n)
{
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to n-1
    for (var i = 2; i < n; i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function for finding number of subsequences
// which consists exactly K primes
function countSubsequences(arr, n, k)
{
    var countPrime = 0;
    for (var i = 0; i < n; i++) {
        if (isPrime(arr[i]))
            countPrime++;
    }
    // if number of primes are less than k
    if (countPrime < k)
        return 0;
 
    return nCr(countPrime, k)
           * Math.pow(2, (n - countPrime));
}
 
// Driver code
var arr = [ 1, 2, 3, 4, 5, 6, 7 ];
var K = 3;
var n = arr.length;
document.write( countSubsequences(arr, n, K));
 
</script>

                    

Output: 
32

 


Time Complexity: O(N^2) . 
 Auxiliary Space: O(1) . 
 



Last Updated : 31 Aug, 2022
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