Check if a number exists having exactly N factors and K prime factors

Difficulty Level : Expert
Last Updated : 12 Apr, 2021

Given two numbers N and K, the task is to find whether an integer X exists such that it has exactly N factors and K out of them are prime.
Examples:

Input: N = 4, K = 2
Output: Yes
Explanation:
One possible number for X is 6.
The number 6 has a total 4 factors: 1, 2, 3 & 6.
It also has exactly 2 prime factors: 2 & 3.
Input: N = 3, K = 1
Output: Yes
Explanation:
One possible number for X is 49.
The number 49 has a total 3 factors: 1, 7, & 49.
It also has exactly 1 prime factor: 7.

Approach: The idea is to use the following identity.

For any number X, if the number has N factors out of which K are prime:

X = k1^a + k2^b + k3^c + ... + knⁿ

The total number of factors N is equal to:

N = (a + 1) * (b + 1) * (c + 1) .. (n + 1)

Therefore, the idea is to check if N can be represented as a product of K integers greater than 1. This can be done by finding the divisors of the number N.
If the count of this is less than K, then the answer is not possible. Else, it is possible.

Below is the implementation of the above approach:

C++

// C++ program to check if it
// is possible to make a number
// having total N factors and
// K prime factors
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to compute the
// number of factors of
// the given number
bool factors(int n, int k)
{
    // Vector to store
    // the prime factors
    vector<int> v;
 
    // While there are no
    // two multiples in
    // the number, divide
    // it by 2
    while (n % 2 == 0) {
        v.push_back(2);
        n /= 2;
    }
 
    // If the size is already
    // greater than K,
    // then return true
    if (v.size() >= k)
        return true;
 
    // Computing the remaining
    // divisors of the number
    for (int i = 3; i * i <= n;
         i += 2) {
 
        // If n is divisible by i,
        // then it is a divisor
        while (n % i == 0) {
            n = n / i;
            v.push_back(i);
        }
 
        // If the size is already
        // greater than K, then
        // return true
        if (v.size() >= k)
            return true;
    }
 
    if (n > 2)
        v.push_back(n);
 
    // If the size is already
    // greater than K,
    // then return true
    if (v.size() >= k)
        return true;
 
    // If none of the above
    // conditions satisfies,
    // then return false
    return false;
}
 
// Function to check if it is
// possible to make a number
// having total N factors and
// K prime factors
void operation(int n, int k)
{
    bool answered = false;
 
    // If total divisors are
    // less than the number
    // of prime divisors,
    // then print No
    if (n < k) {
        answered = true;
        cout << "No"
             << "\n";
    }
 
    // Find the number of
    // factors of n
    bool ok = factors(n, k);
 
    if (!ok && !answered) {
        answered = true;
        cout << "No"
             << "\n";
    }
 
    if (ok && !answered)
        cout << "Yes"
             << "\n";
}
 
// Driver Code
int main()
{
    int n = 4;
    int k = 2;
 
    operation(n, k);
    return 0;
}

Java

// Java program to check if it
// is possible to make a number
// having total N factors and
// K prime factors
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to compute the
// number of factors of
// the given number
static boolean factors(int n, int k)
{
     
    // Vector to store
    // the prime factors
    ArrayList<Integer> v = new ArrayList<Integer>();
 
    // While there are no
    // two multiples in
    // the number, divide
    // it by 2
    while (n % 2 == 0)
    {
        v.add(2);
        n /= 2;
    }
 
    // If the size is already
    // greater than K,
    // then return true
    if (v.size() >= k)
        return true;
 
    // Computing the remaining
    // divisors of the number
    for(int i = 3; i * i <= n; i += 2)
    {
        
       // If n is divisible by i,
       // then it is a divisor
       while (n % i == 0)
       {
           n = n / i;
           v.add(i);
       }
        
       // If the size is already
       // greater than K, then
       // return true
       if (v.size() >= k)
           return true;
    }
 
    if (n > 2)
        v.add(n);
 
    // If the size is already
    // greater than K,
    // then return true
    if (v.size() >= k)
        return true;
 
    // If none of the above
    // conditions satisfies,
    // then return false
    return false;
}
     
// Function to check if it is
// possible to make a number
// having total N factors and
// K prime factors
static void operation(int n, int k)
{
    boolean answered = false;
 
    // If total divisors are
    // less than the number
    // of prime divisors,
    // then print No
    if (n < k)
    {
        answered = true;
        System.out.println("No");
    }
 
    // Find the number of
    // factors of n
    boolean ok = factors(n, k);
 
    if (!ok && !answered)
    {
        answered = true;
        System.out.println("No");
    }
    if (ok && !answered)
        System.out.println("Yes");
}
     
// Driver code
public static void main(String[] args)
{
    int n = 4;
    int k = 2;
     
    //Function call
    operation(n, k);
}
}
 
// This code is contributed by coder001

Python3

# Python3 program to check if it
# is possible to make a number
# having total N factors and
# K prime factors
 
# Function to compute the
# number of factors of
# the given number
def factors(n, k):
 
    # Vector to store
    # the prime factors
    v = [];
 
    # While there are no
    # two multiples in
    # the number, divide
    # it by 2
    while (n % 2 == 0):
        v.append(2);
        n //= 2;
     
    # If the size is already
    # greater than K,
    # then return true
    if (len(v) >= k):
        return True;
 
    # Computing the remaining
    # divisors of the number
    for i in range(3, int(n ** (1 / 2)), 2):
 
        # If n is divisible by i,
        # then it is a divisor
        while (n % i == 0):
            n = n // i;
            v.append(i);
 
        # If the size is already
        # greater than K, then
        # return true
        if (len(v) >= k):
            return True;
     
    if (n > 2):
        v.append(n);
 
    # If the size is already
    # greater than K,
    # then return true
    if (len(v) >= k):
        return True;
 
    # If none of the above
    # conditions satisfies,
    # then return false
    return False;
 
# Function to check if it is
# possible to make a number
# having total N factors and
# K prime factors
def operation(n, k):
    answered = False;
 
    # If total divisors are
    # less than the number
    # of prime divisors,
    # then print No
    if (n < k):
        answered = True;
        print("No");
 
    # Find the number of
    # factors of n
    ok = factors(n, k);
 
    if (not ok and not answered):
        answered = True;
        print("No");
 
    if (ok and not answered):
        print("Yes");
 
# Driver Code
if __name__ == "__main__":
 
    n = 4;
    k = 2;
    operation(n, k);
 
# This code is contributed by AnkitRai01

C#

// C# program to check if it
// is possible to make a number
// having total N factors and
// K prime factors
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to compute the
// number of factors of
// the given number
static bool factors(int n, int k)
{
     
    // Vector to store
    // the prime factors
    List<int> v = new List<int>();
 
    // While there are no
    // two multiples in
    // the number, divide
    // it by 2
    while (n % 2 == 0)
    {
        v.Add(2);
        n /= 2;
    }
 
    // If the size is already
    // greater than K,
    // then return true
    if (v.Count >= k)
        return true;
 
    // Computing the remaining
    // divisors of the number
    for(int i = 3; i * i <= n; i += 2)
    {
         
        // If n is divisible by i,
        // then it is a divisor
        while (n % i == 0)
        {
            n = n / i;
            v.Add(i);
        }
             
        // If the size is already
        // greater than K, then
        // return true
        if (v.Count >= k)
            return true;
    }
 
    if (n > 2)
        v.Add(n);
 
    // If the size is already
    // greater than K,
    // then return true
    if (v.Count >= k)
        return true;
 
    // If none of the above
    // conditions satisfies,
    // then return false
    return false;
}
     
// Function to check if it is
// possible to make a number
// having total N factors and
// K prime factors
static void operation(int n, int k)
{
    bool answered = false;
 
    // If total divisors are
    // less than the number
    // of prime divisors,
    // then print No
    if (n < k)
    {
        answered = true;
        Console.WriteLine("No");
    }
 
    // Find the number of
    // factors of n
    bool ok = factors(n, k);
 
    if (!ok && !answered)
    {
        answered = true;
        Console.WriteLine("No");
    }
    if (ok && !answered)
        Console.WriteLine("Yes");
}
     
// Driver code
public static void Main()
{
    int n = 4;
    int k = 2;
     
    // Function call
    operation(n, k);
}
}
 
// This code is contributed by sanjoy_62

Javascript

<script>
 
// Javascript program to check if it
// is possible to make a number
// having total N factors and
// K prime factors
 
// Function to compute the
// number of factors of
// the given number
function factors(n, k)
{
    // Vector to store
    // the prime factors
    let v = [];
 
    // While there are no
    // two multiples in
    // the number, divide
    // it by 2
    while (n % 2 == 0) {
        v.push(2);
        n = parseInt(n/2);
    }
 
    // If the size is already
    // greater than K,
    // then return true
    if (v.length >= k)
        return true;
 
    // Computing the remaining
    // divisors of the number
    for (let i = 3; i * i <= n;
         i += 2) {
 
        // If n is divisible by i,
        // then it is a divisor
        while (n % i == 0) {
            n = parseInt(n / i);
            v.push(i);
        }
 
        // If the size is already
        // greater than K, then
        // return true
        if (v.length >= k)
            return true;
    }
 
    if (n > 2)
        v.push(n);
 
    // If the size is already
    // greater than K,
    // then return true
    if (v.length >= k)
        return true;
 
    // If none of the above
    // conditions satisfies,
    // then return false
    return false;
}
 
// Function to check if it is
// possible to make a number
// having total N factors and
// K prime factors
function operation(n, k)
{
    let answered = false;
 
    // If total divisors are
    // less than the number
    // of prime divisors,
    // then print No
    if (n < k) {
        answered = true;
        document.write("No<br>");
    }
 
    // Find the number of
    // factors of n
    let ok = factors(n, k);
 
    if (!ok && !answered) {
        answered = true;
        document.write("No<br>");
    }
 
    if (ok && !answered)
        document.write("Yes<br>");
}
 
// Driver Code
let n = 4;
let k = 2;
 
operation(n, k);
 
</script>

Output:

Yes

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