Check if there exists a number with X factors out of which exactly K are prime

Given two integers X and K, the task is to determine whether there exists a number that has exactly X factors out of which K is prime.

Examples:

Input: X = 8, K = 1
Output: Yes
Explanation:
The number is 128
Factors of 128 = {1, 2, 4, 8, 16, 32, 64, 128} which are 8 in count = X
Among these, only 2 is prime. Therefore count of prime factor = 1 = K

Input: X = 4, K = 2
Output: Yes
Explanation:
The number is 6
Factors of 6 = {1, 2, 3, 6} which are 4 in count = X
Among these, only 2 and 3 are prime. Therefore count of prime factor = 2 = K

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Suppose a number N has X factors out of which K are prime, say $K_{1}, K_{2}, K_{3}, ...K_{M}$
Thus, number can be written as $N = K_{1}^{a}, K_{2}^{b}, K_{3}^{c}, ...K_{M}^{K}$ where, the total number of factors is calculated by $X = (a+1) * (b+1) * (c+1) * ... *(K+1)$
It is observed that X is a product of “power+1” of the prime factors of the number. Thus, if we are able to divide X into a product of K numbers, then we can form a number with exactly X factors out of which K is prime.

Below is the implementation of the above approach:

C++

// C++ program to check if there exists 
// a number with X factors 
// out of which exactly K are prime 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to check if such number exists 
bool check(int X, int K) 
{ 
    int prime, temp, sqr, i; 
  
    // To store the sum of powers 
    // of prime factors of X which 
    // determines the maximum count 
    // of numbers whose product can form X 
    prime = 0; 
    temp = X; 
    sqr = sqrt(X); 
  
    // Determining the prime factors of X 
    for (i = 2; i <= sqr; i++) { 
  
        while (temp % i == 0) { 
            temp = temp / i; 
            prime++; 
        } 
    } 
  
    // To check if the number is prime 
    if (temp > 2) 
        prime++; 
  
    // If X is 1, then we cannot form 
    // a number with 1 factor and K 
    // prime factor (as K is atleast 1) 
    if (X == 1) 
        return false; 
  
    // If X itself is prime then it 
    // can be represented as a power 
    // of only 1 prime factor which 
    // is X itself so we return true 
    if (prime == 1 && K == 1) 
        return true; 
  
    // If sum of the powers of prime factors 
    // of X is greater than or equal to K, 
    // which means X can be represented as a 
    // product of K numbers, we return true 
    else if (prime >= K) 
        return true; 
  
    // In any other case, we return false 
    // as we cannot form a number with X 
    // factors and K prime factors 
    else
        return false; 
} 
  
// Driver code 
int main() 
{ 
    int X, K; 
    X = 4; 
    K = 2; 
  
    if (check(X, K)) 
        cout << "Yes"; 
    else
        cout << "No"; 
} 

Java

// Java program to check if there exists 
// a number with X factors 
// out of which exactly K are prime 
   
  
import java.util.*; 
  
class GFG{ 
   
// Function to check if such number exists 
static boolean check(int X, int K) 
{ 
    int prime, temp, sqr, i; 
   
    // To store the sum of powers 
    // of prime factors of X which 
    // determines the maximum count 
    // of numbers whose product can form X 
    prime = 0; 
    temp = X; 
    sqr = (int) Math.sqrt(X); 
   
    // Determining the prime factors of X 
    for (i = 2; i <= sqr; i++) { 
   
        while (temp % i == 0) { 
            temp = temp / i; 
            prime++; 
        } 
    } 
   
    // To check if the number is prime 
    if (temp > 2) 
        prime++; 
   
    // If X is 1, then we cannot form 
    // a number with 1 factor and K 
    // prime factor (as K is atleast 1) 
    if (X == 1) 
        return false; 
   
    // If X itself is prime then it 
    // can be represented as a power 
    // of only 1 prime factor which 
    // is X itself so we return true 
    if (prime == 1 && K == 1) 
        return true; 
   
    // If sum of the powers of prime factors 
    // of X is greater than or equal to K, 
    // which means X can be represented as a 
    // product of K numbers, we return true 
    else if (prime >= K) 
        return true; 
   
    // In any other case, we return false 
    // as we cannot form a number with X 
    // factors and K prime factors 
    else
        return false; 
} 
   
// Driver code 
public static void main(String[] args) 
{ 
    int X, K; 
    X = 4; 
    K = 2; 
   
    if (check(X, K)) 
        System.out.print("Yes"); 
    else
        System.out.print("No"); 
} 
} 
  
// This code contributed by Rajput-Ji 

Python3

# Python3 program to check if there exists 
# a number with X factors 
# out of which exactly K are prime 
  
from math import sqrt 
# Function to check if such number exists 
def check(X,K): 
  
    # To store the sum of powers 
    # of prime factors of X which 
    # determines the maximum count 
    # of numbers whose product can form X 
    prime = 0
    temp = X 
    sqr = int(sqrt(X)) 
  
    # Determining the prime factors of X 
    for i in range(2,sqr+1,1): 
        while (temp % i == 0): 
            temp = temp // i 
            prime += 1
  
    # To check if the number is prime 
    if (temp > 2): 
        prime += 1
  
    # If X is 1, then we cannot form 
    # a number with 1 factor and K 
    # prime factor (as K is atleast 1) 
    if (X == 1): 
        return False
  
    # If X itself is prime then it 
    # can be represented as a power 
    # of only 1 prime factor w0hich 
    # is X itself so we return true 
    if (prime == 1 and K == 1): 
        return True
  
    # If sum of the powers of prime factors 
    # of X is greater than or equal to K, 
    # which means X can be represented as a 
    # product of K numbers, we return true 
    elif(prime >= K): 
        return True
  
    # In any other case, we return false 
    # as we cannot form a number with X 
    # factors and K prime factors 
    else: 
        return False
  
# Driver code 
if __name__ == '__main__': 
    X = 4
    K = 2
  
    if (check(X, K)): 
        print("Yes") 
    else: 
        print("No") 
  
# This code is contributed by Surendra_Gangwar 

C#

// C# program to check if there exists 
// a number with X factors 
// out of which exactly K are prime 
using System; 
  
class GFG{ 
  
    // Function to check if such number exists 
    static bool check(int X, int K) 
    { 
        int prime, temp, sqr, i; 
      
        // To store the sum of powers 
        // of prime factors of X which 
        // determines the maximum count 
        // of numbers whose product can form X 
        prime = 0; 
        temp = X; 
        sqr = Convert.ToInt32(Math.Sqrt(X)); 
      
        // Determining the prime factors of X 
        for (i = 2; i <= sqr; i++) { 
      
            while (temp % i == 0) { 
                temp = temp / i; 
                prime++; 
            } 
        } 
      
        // To check if the number is prime 
        if (temp > 2) 
            prime++; 
      
        // If X is 1, then we cannot form 
        // a number with 1 factor and K 
        // prime factor (as K is atleast 1) 
        if (X == 1) 
            return false; 
      
        // If X itself is prime then it 
        // can be represented as a power 
        // of only 1 prime factor which 
        // is X itself so we return true 
        if (prime == 1 && K == 1) 
            return true; 
      
        // If sum of the powers of prime factors 
        // of X is greater than or equal to K, 
        // which means X can be represented as a 
        // product of K numbers, we return true 
        else if (prime >= K) 
            return true; 
      
        // In any other case, we return false 
        // as we cannot form a number with X 
        // factors and K prime factors 
        else
            return false; 
    } 
      
    // Driver code 
    static public void Main () 
    { 
        int X, K; 
        X = 4; 
        K = 2; 
      
        if (check(X, K)) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    } 
} 
  
// This code is contributed by shubhamsingh10 

Output:

Yes

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

Recommended Posts: