Check if a number can be expressed as a product of exactly K prime divisors

Last Updated : 12 Jul, 2025

Given an integer N, the task is to check if it can be expressed as a product of exactly K prime divisors.
Examples:

Input: N = 12, K = 3
Output: Yes
Explanation:
12 can be expressed as product of 2×2×3.

Input: N = 14, K = 3
Output:  No
Explanation:
14 can be only expressed as product of 2×7.

Approach:
To solve the problem mentioned above we are given the value N and we will find the maximum number of values we can split N into. We can represent prime factorization of N as \prod_{i=1}^{K} {p_{i}}^{a_{i}} where p_i are the prime factors of N and a_i are the exponents. We know that total number of divisors of N is \prod_{i=1}^{K} (a_{i}+1) . Therefore, we can observe that we have to check whether it is possible to represent N as product of K numbers or not. If the maximum split is less than K then it is not possible to express it in exactly K prime divisors, else it is always possible.

C++

// CPP implementation to Check if a
// number can be expressed as a
// product of exactly K prime divisors

#include <bits/stdc++.h>
using namespace std;

// function to find K prime divisors
void KPrimeDivisors(int N, int K)
{
    int maximum_split = 0;

    // count number of 2s that divide N
    while (N % 2 == 0) {
        maximum_split++;
        N /= 2;
    }

    // N must be odd at this point.
    // So we can skip one element
    for (int i = 3; i * i <= N; i = i + 2) {

        while (N % i == 0) {
            // divide the value of N
            N = N / i;

            // increment count
            maximum_split++;
        }
    }

    // Condition to handle the case when n
    // is a prime number greater than 2
    if (N > 2)
        maximum_split++;

    // check if maximum_split is less than K
    // then it not possible
    if (maximum_split < K) {
        printf("No\n");
        return;
    }

    printf("Yes\n");
}

/* Driver code */
int main()
{
    // initialise N and K
    int N = 12;
    int K = 3;

    KPrimeDivisors(N, K);

    return 0;
}

Java

// Java implementation to Check if a 
// number can be expressed as a 
// product of exactly K prime divisors 
class GFG {
    
    // function to find K prime divisors 
    static void KPrimeDivisors(int N, int K) 
    { 
        int maximum_split = 0; 
    
        // count number of 2s that divide N 
        while (N % 2 == 0) { 
            maximum_split++; 
            N /= 2; 
        } 
    
        // N must be odd at this point. 
        // So we can skip one element 
        for (int i = 3; i * i <= N; i = i + 2) { 
    
            while (N % i == 0) { 
                // divide the value of N 
                N = N / i; 
    
                // increment count 
                maximum_split++; 
            } 
        } 
    
        // Condition to handle the case when n 
        // is a prime number greater than 2 
        if (N > 2) 
            maximum_split++; 
    
        // check if maximum_split is less than K 
        // then it not possible 
        if (maximum_split < K) { 
            System.out.println("No"); 
            return; 
        } 
    
        System.out.println("Yes"); 
    } 
    
    /* Driver code */
    public static void main (String[] args)
    { 
        // initialise N and K 
        int N = 12; 
        int K = 3; 
    
        KPrimeDivisors(N, K); 
    } 
}

// This code is contributed by Yash_R

Python3

# Python implementation to Check if a
# number can be expressed as a
# product of exactly K prime divisors

import math as mt 

# function to find K prime divisors
def KPrimeDivisors(n, k): 
    
    # To count maximum split of N
    maximum_split = 0
    
    # count number of 2s that divide N
    while n % 2 == 0: 
        maximum_split+= 1
        n = n // 2
        
    # n must be odd at this point 
    # so we skip one element 
    for i in range(3, mt.ceil(mt.sqrt(n)), 2): 
        while n % i == 0: 
            n = n / i; 
            maximum_split+= 1
            
    # Condition to handle the case when n
    # is a prime number greater than 2 
    if n > 2: 
        maximum_split+= 1
        
    # check if maximum_split is less than K
    # then it not possible
    if maximum_split < k: 
        print("No")
        return

    print("Yes")
        
    

# Driver code 
N = 12
K = 3
KPrimeDivisors(N, K)

    
// C# implementation to Check if a 
// number can be expressed as a 
// product of exactly K prime divisors 
using System;

class GFG {
     
    // function to find K prime divisors 
    static void KPrimeDivisors(int N, int K) 
    { 
        int maximum_split = 0; 
     
        // count number of 2s that divide N 
        while (N % 2 == 0) { 
            maximum_split++; 
            N /= 2; 
        } 
     
        // N must be odd at this point. 
        // So we can skip one element 
        for (int i = 3; i * i <= N; i = i + 2) { 
     
            while (N % i == 0) { 

                // divide the value of N 
                N = N / i; 
     
                // increment count 
                maximum_split++; 
            } 
        } 
     
        // Condition to handle the case when n 
        // is a prime number greater than 2 
        if (N > 2) 
            maximum_split++; 
     
        // check if maximum_split is less than K 
        // then it not possible 
        if (maximum_split < K) { 
            Console.WriteLine("No"); 
            return; 
        } 
     
        Console.WriteLine("Yes"); 
    } 
     
    /* Driver code */
    public static void Main(String[] args)
    { 
        // initialise N and K 
        int N = 12; 
        int K = 3; 
     
        KPrimeDivisors(N, K); 
    } 
}

// This code is contributed by 29AjayKumar

JavaScript

<script>
// javascript implementation to Check if a 
// number can be expressed as a 
// product of exactly K prime divisors     

// function to find K prime divisors
    function KPrimeDivisors(N , K)
    {
        var maximum_split = 0;

        // count number of 2s that divide N
        while (N % 2 == 0)
        {
            maximum_split++;
            N /= 2;
        }

        // N must be odd at this point.
        // So we can skip one element
        for (i = 3; i * i <= N; i = i + 2) 
        {

            while (N % i == 0)
            {
            
                // divide the value of N
                N = N / i;

                // increment count
                maximum_split++;
            }
        }

        // Condition to handle the case when n
        // is a prime number greater than 2
        if (N > 2)
            maximum_split++;

        // check if maximum_split is less than K
        // then it not possible
        if (maximum_split < K)
        {
            document.write("No");
            return;
        }

        document.write("Yes");
    }

    /* Driver code */
    
        // initialise N and K
        var N = 12;
        var K = 3;
        KPrimeDivisors(N, K);

// This code is contributed by gauravrajput1. 
</script>

Output:

Yes

Time Complexity: O(sqrt(N))

Auxiliary Space: O(1)

Analysis of Algorithms

saurabhshadow

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