Generating all divisors of a number using its prime factorization

Given an integer N, the task is to find all of its divisors using its prime factorization.

Examples:

Input: N = 6 
Output: 1 2 3 6

Input: N = 10 
Output: 1 2 5 10

Approach: As every number greater than 1 can be represented in its prime factorization as p₁^a₁*p₂^a₂*……*p_k^a_k, where p_i is a prime number, k ≥ 1 and a_i is a positive integer. 
Now all the possible divisors can be generated recursively if the count of occurrence of every prime factor of n is known. For every prime factor p_i, it can be included x times where 0 ≤ x ≤ a_i. First, find the prime factorization of n using this approach and for every prime factor, store it with the count of its occurrence.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include "iostream"
#include "vector"
using namespace std;

struct primeFactorization {

    // to store the prime factor
    // and its highest power
    int countOfPf, primeFactor;
};

// Recursive function to generate all the
// divisors from the prime factors
void generateDivisors(int curIndex, int curDivisor,
                      vector<primeFactorization>& arr)
{

    // Base case i.e. we do not have more
    // primeFactors to include
    if (curIndex == arr.size()) {
        cout << curDivisor << ' ';
        return;
    }

    for (int i = 0; i <= arr[curIndex].countOfPf; ++i) {
        generateDivisors(curIndex + 1, curDivisor, arr);
        curDivisor *= arr[curIndex].primeFactor;
    }
}

// Function to find the divisors of n
void findDivisors(int n)
{

    // To store the prime factors along
    // with their highest power
    vector<primeFactorization> arr;

    // Finding prime factorization of n
    for (int i = 2; i * i <= n; ++i) {
        if (n % i == 0) {
            int count = 0;
            while (n % i == 0) {
                n /= i;
                count += 1;
            }

            // For every prime factor we are storing
            // count of it's occurrenceand itself.
            arr.push_back({ count, i });
        }
    }

    // If n is prime
    if (n > 1) {
        arr.push_back({ 1, n });
    }

    int curIndex = 0, curDivisor = 1;

    // Generate all the divisors
    generateDivisors(curIndex, curDivisor, arr);
}

// Driver code
int main()
{
    int n = 6;

    findDivisors(n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
class GFG
{

static class primeFactorization
{

    // to store the prime factor
    // and its highest power
    int countOfPf, primeFactor;

    public primeFactorization(int countOfPf,
                              int primeFactor)
    {
        this.countOfPf = countOfPf;
        this.primeFactor = primeFactor;
    }
}

// Recursive function to generate all the
// divisors from the prime factors
static void generateDivisors(int curIndex, int curDivisor,
                           Vector<primeFactorization> arr)
{

    // Base case i.e. we do not have more
    // primeFactors to include
    if (curIndex == arr.size())
    {
        System.out.print(curDivisor + " ");
        return;
    }

    for (int i = 0; i <= arr.get(curIndex).countOfPf; ++i)
    {
        generateDivisors(curIndex + 1, curDivisor, arr);
        curDivisor *= arr.get(curIndex).primeFactor;
    }
}

// Function to find the divisors of n
static void findDivisors(int n)
{

    // To store the prime factors along
    // with their highest power
    Vector<primeFactorization> arr = new Vector<>();

    // Finding prime factorization of n
    for (int i = 2; i * i <= n; ++i)
    {
        if (n % i == 0)
        {
            int count = 0;
            while (n % i == 0)
            {
                n /= i;
                count += 1;
            }

            // For every prime factor we are storing
            // count of it's occurrenceand itself.
            arr.add(new primeFactorization(count, i ));
        }
    }

    // If n is prime
    if (n > 1)
    {
        arr.add(new primeFactorization( 1, n ));
    }

    int curIndex = 0, curDivisor = 1;

    // Generate all the divisors
    generateDivisors(curIndex, curDivisor, arr);
}

// Driver code
public static void main(String []args)
{
    int n = 6;

    findDivisors(n);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach

# Recursive function to generate all the
# divisors from the prime factors
def generateDivisors(curIndex, curDivisor, arr):

    # Base case i.e. we do not have more
    # primeFactors to include
    if (curIndex == len(arr)):
        print(curDivisor, end = ' ')
        return

    for i in range(arr[curIndex][0] + 1):
        generateDivisors(curIndex + 1, curDivisor, arr)
        curDivisor *= arr[curIndex][1]

# Function to find the divisors of n
def findDivisors(n):

    # To store the prime factors along
    # with their highest power
    arr = []

    # Finding prime factorization of n
    i = 2
    while(i * i <= n):
        if (n % i == 0):
            count = 0
            while (n % i == 0):
                n //= i
                count += 1

            # For every prime factor we are storing
            # count of it's occurrenceand itself.
            arr.append([count, i])

    # If n is prime
    if (n > 1):
        arr.append([1, n])

    curIndex = 0
    curDivisor = 1

    # Generate all the divisors
    generateDivisors(curIndex, curDivisor, arr)

# Driver code
n = 6
findDivisors(n)

# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

public class primeFactorization
{

    // to store the prime factor
    // and its highest power
    public int countOfPf, primeFactor;

    public primeFactorization(int countOfPf,
                              int primeFactor)
    {
        this.countOfPf = countOfPf;
        this.primeFactor = primeFactor;
    }
}

// Recursive function to generate all the
// divisors from the prime factors
static void generateDivisors(int curIndex, int curDivisor,
                             List<primeFactorization> arr)
{

    // Base case i.e. we do not have more
    // primeFactors to include
    if (curIndex == arr.Count)
    {
        Console.Write(curDivisor + " ");
        return;
    }

    for (int i = 0; i <= arr[curIndex].countOfPf; ++i)
    {
        generateDivisors(curIndex + 1, curDivisor, arr);
        curDivisor *= arr[curIndex].primeFactor;
    }
}

// Function to find the divisors of n
static void findDivisors(int n)
{

    // To store the prime factors along
    // with their highest power
    List<primeFactorization> arr = new List<primeFactorization>();

    // Finding prime factorization of n
    for (int i = 2; i * i <= n; ++i)
    {
        if (n % i == 0)
        {
            int count = 0;
            while (n % i == 0)
            {
                n /= i;
                count += 1;
            }

            // For every prime factor we are storing
            // count of it's occurrenceand itself.
            arr.Add(new primeFactorization(count, i ));
        }
    }

    // If n is prime
    if (n > 1)
    {
        arr.Add(new primeFactorization( 1, n ));
    }

    int curIndex = 0, curDivisor = 1;

    // Generate all the divisors
    generateDivisors(curIndex, curDivisor, arr);
}

// Driver code
public static void Main(String []args)
{
    int n = 6;

    findDivisors(n);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript implementation of the approach

// Recursive function to generate all the
// divisors from the prime factors
function generateDivisors(curIndex, curDivisor, arr)
{

    // Base case i.e. we do not have more
    // primeFactors to include
    if (curIndex == arr.length) {
        document.write(curDivisor + " ");
        return;
    }

    for (var i = 0; i <= arr[curIndex][0]; ++i) {
        generateDivisors(curIndex + 1, curDivisor, arr);
        curDivisor *= arr[curIndex][1];
    }
}

// Function to find the divisors of n
function findDivisors(n)
{

    // To store the prime factors along
    // with their highest power
    arr = [];

    // Finding prime factorization of n
    for (var i = 2; i * i <= n; ++i) {
        if (n % i == 0) {
            var count = 0;
            while (n % i == 0) {
                n /= i;
                count += 1;
            }

            // For every prime factor we are storing
            // count of it's occurrenceand itself.
            arr.push([ count, i ]);
        }
    }

    // If n is prime
    if (n > 1) {
        arr.push([ 1, n ]);
    }

    var curIndex = 0, curDivisor = 1;

    // Generate all the divisors
    generateDivisors(curIndex, curDivisor, arr);
}


// driver code
var n = 6;
findDivisors(n);
// This code contributed by shubhamsingh10
</script>

Output:

1 3 2 6

Time Complexity: O(sqrt(n))
Auxiliary Space: O(sqrt(n))