Print all combinations of factors (Ways to factorize)

Write a program to print all the combinations of factors of given number n.

Examples:

Input : 16
Output :2 2 2 2 
        2 2 4 
        2 8 
        4 4 

Input : 12
Output : 2 2 3
         2 6
         3 4

To solve this problem we take one array of array of integers or list of list of integers to store all the factors combination possible for the given n. So, to achieve this we can have one recursive function which can store the factors combination in each of its iteration. And each of those list should be stored in the final result list.

Below is the implementation of the above approach.

C++

#include <iostream>
#include <vector>

using namespace std;

// Function to find all factor combinations of a number
void backtrack(int start, int target, vector<int>& factors,
               vector<vector<int> >& combinations, int n)
{
    // Base case: if target is 1, we have found a
    // factorization
    if (target == 1) {
        // Add a copy of the factors vector to combinations,
        // except if the vector contains only one factor
        // equal to n
        if (factors.size() > 1 || factors[0] != n) {
            combinations.push_back(factors);
        }
        return;
    }

    // Try all factors from start to target, inclusive
    for (int i = start; i <= target; i++) {
        if (target % i == 0) {
            factors.push_back(i); // Add i to factors
            backtrack(i, target / i, factors, combinations,
                      n); // Recursively find factors of
                          // target / i
            factors.pop_back(); // Remove i from factors
        }
    }
}

// Function to call backtrack and return the factor
// combinations
vector<vector<int> > factorCombinations(int n)
{
    vector<vector<int> > combinations;
    vector<int> factors;
    backtrack(2, n, factors, combinations, n);
    return combinations;
}

int main()
{
    int n = 12;
    vector<vector<int> > combinations
        = factorCombinations(n);

    cout << "All the combinations of factors of " << n
         << " are:" << endl;
    for (vector<int> combination : combinations) {
        for (int factor : combination) {
            cout << factor << " ";
        }
        cout << endl;
    }

    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

public class FactorCombinations {

    public static List<List<Integer> >
    factorCombinations(int n)
    {
        List<List<Integer> > combinations
            = new ArrayList<>();

        backtrack(2, n, new ArrayList<>(), combinations, n);

        return combinations;
    }

    private static void
    backtrack(int start, int target, List<Integer> factors,
              List<List<Integer> > combinations, int n)
    {
        if (target == 1) {
            // add a copy of the factors list to
            // combinations, except if the list contains
            // only one factor equal to n
            if (factors.size() > 1 || factors.get(0) != n) {
                combinations.add(new ArrayList<>(factors));
            }
            return;
        }

        // try all factors from start to target, inclusive
        for (int i = start; i <= target; i++) {
            if (target % i == 0) {
                factors.add(i);
                backtrack(i, target / i, factors,
                          combinations, n);
                factors.remove(factors.size() - 1);
            }
        }
    }

    public static void main(String[] args)
    {
        int n = 12;
        List<List<Integer> > combinations
            = factorCombinations(n);

        System.out.printf(
            "All the combinations of factors of %d are:%n",
            n);
        for (List<Integer> combination : combinations) {
            System.out.println(combination);
        }
    }
}

Python

def backtrack(start, target, factors, combinations, n):
    # Base case: if target is 1, we have found a factorization
    if target == 1:
        # Add a copy of the factors list to combinations, except if the list contains only one factor equal to n
        if len(factors) > 1 or factors[0] != n:
            combinations.append(list(factors))
        return

    # Try all factors from start to target, inclusive
    for i in range(start, target + 1):
        if target % i == 0:
            factors.append(i)  # Add i to factors
            # Recursively find factors of target // i
            backtrack(i, target // i, factors, combinations, n)
            factors.pop()  # Remove i from factors


def factorCombinations(n):
    combinations = []
    factors = []
    backtrack(2, n, factors, combinations, n)
    return combinations


if __name__ == '__main__':
    n = 12
    combinations = factorCombinations(n)
    print("All the combinations of factors of {} are:".format(n))
    for combination in combinations:
        print(combination)

C#

using System;
using System.Collections.Generic;

namespace FactorCombinations {
class Program {
    static void Main(string[] args)
    {
        int n = 12;
        List<List<int> > combinations
            = FactorCombinations(n);

        Console.WriteLine(
            $
            "All the combinations of factors of {n} are:");
        foreach(List<int> combination in combinations)
        {
            Console.WriteLine(
                string.Join(" ", combination));
        }
    }

    static List<List<int> > FactorCombinations(int n)
    {
        List<List<int> > combinations
            = new List<List<int> >();

        Backtrack(2, n, new List<int>(), combinations, n);

        return combinations;
    }

    static void Backtrack(int start, int target,
                          List<int> factors,
                          List<List<int> > combinations,
                          int n)
    {
        if (target == 1) {
            // add a copy of the factors list to
            // combinations, except if the list contains
            // only one factor equal to n
            if (factors.Count > 1 || factors[0] != n) {
                combinations.Add(new List<int>(factors));
            }
            return;
        }

        // try all factors from start to target, inclusive
        for (int i = start; i <= target; i++) {
            if (target % i == 0) {
                factors.Add(i);
                Backtrack(i, target / i, factors,
                          combinations, n);
                factors.RemoveAt(factors.Count - 1);
            }
        }
    }
}
}

Javascript

// Function to find all factor combinations of a number
function backtrack(start, target, factors, combinations, n) {
    // Base case: if target is 1, we have found a factorization
    if (target === 1) {
        // Add a copy of the factors array to combinations, except if the array contains only one factor equal to n
        if (factors.length > 1 || factors[0] !== n) {
            combinations.push([...factors]);
        }
        return;
    }

    // Try all factors from start to target, inclusive
    for (let i = start; i <= target; i++) {
        if (target % i === 0) {
            factors.push(i);  // Add i to factors
            backtrack(i, target / i, factors, combinations, n);  // Recursively find factors of target / i
            factors.pop();  // Remove i from factors
        }
    }
}

// Function to call backtrack and return the factor combinations
function factorCombinations(n) {
    const combinations = [];
    const factors = [];
    backtrack(2, n, factors, combinations, n);
    return combinations;
}

// Main code
const n = 12;
const combinations = factorCombinations(n);

console.log(`All the combinations of factors of ${n} are:`);
for (let combination of combinations) {
    console.log(combination.join(' '));
}

Output

All the combinations of factors of 12 are:
[2, 2, 3]
[2, 6]
[3, 4]

Time Complexity:  O(sqrt(n) * log(n)), where n is the input integer.  
Auxiliary Space: O(log(n)), where n is the input integer.  

Another Approach:

The code below is pure recursive code for printing all combinations of factors:

It uses a vector of integer to store a single list of factors and a vector of integer to store all combinations of factors. Instead of using an iterative loop, it uses the same recursive function to calculate all factor combinations.

C++

// C++ program to print all factors combination
#include <bits/stdc++.h>
using namespace std;

// vector of vector for storing
// list of factor combinations
vector<vector<int> > factors_combination;

// recursive function
void compute_factors(int current_no, int n, int product,
                     vector<int> single_list)
{

    // base case: if the product
    // exceeds our given number;
    // OR
    // current_no exceeds half the given n
    if (current_no > (n / 2) || product > n)
        return;

    // if current list of factors
    // is contributing to n
    if (product == n) {

        // storing the list
        factors_combination.push_back(single_list);

        // into factors_combination
        return;
    }

    // including current_no in our list
    single_list.push_back(current_no);

    // trying to get required
    // n with including current
    // current_no
    compute_factors(current_no, n, product * current_no,
                    single_list);

    // excluding current_no from our list
    single_list.pop_back();

    // trying to get required n
    // without including current
    // current_no
    compute_factors(current_no + 1, n, product,
                    single_list);
}

// Driver Code
int main()
{
    int n = 16;

    // vector to store single list of factors
    // eg. 2,2,2,2 is one of the list for n=16
    vector<int> single_list;

    // compute_factors ( starting_no, given_n,
    // our_current_product, vector )
    compute_factors(2, n, 1, single_list);

    // printing all possible factors stored in
    // factors_combination
    for (int i = 0; i < factors_combination.size(); i++) {
        for (int j = 0; j < factors_combination[i].size();
             j++)
            cout << factors_combination[i][j] << " ";
        cout << endl;
    }
    return 0;
}

// code contributed by Devendra Kolhe

Java

// Java program to print all factors combination
import java.util.*;

class GFG{

// vector of vector for storing
// list of factor combinations
static Vector<Vector<Integer>> factors_combination =
   new Vector<Vector<Integer>>();

// Recursive function
static void compute_factors(int current_no, int n, int product,
                            Vector<Integer> single_list)
{

    // base case: if the product
    // exceeds our given number;
    // OR
    // current_no exceeds half the given n
    if (current_no > (n / 2) || product > n)
        return;

    // If current list of factors
    // is contributing to n
    if (product == n)
    {

        // Storing the list
        factors_combination.add(single_list);

        // Printing all possible factors stored in
        // factors_combination
        for(int i = 0; i < factors_combination.size(); i++)
        {
            for(int j = 0; j < factors_combination.get(i).size(); j++)
                System.out.print(factors_combination.get(i).get(j) + " ");
        }
        System.out.println();
        factors_combination = new Vector<Vector<Integer>>();

        // Into factors_combination
        return;
    }

    // Including current_no in our list
    single_list.add(current_no);

    // Trying to get required
    // n with including current
    // current_no
    compute_factors(current_no, n,
                    product * current_no,
                    single_list);

    // Excluding current_no from our list
    single_list.remove(single_list.size() - 1);

    // Trying to get required n
    // without including current
    // current_no
    compute_factors(current_no + 1, n, product,
                    single_list);
}

// Driver code
public static void main(String[] args)
{
    int n = 16;

    // Vector to store single list of factors
    // eg. 2,2,2,2 is one of the list for n=16
    Vector<Integer> single_list = new Vector<Integer>();

    // compute_factors ( starting_no, given_n,
    // our_current_product, vector )
    compute_factors(2, n, 1, single_list);
}
}

// This code is contributed by decode2207

Python3

# Python3 program to print all factors combination

# vector of vector for storing
# list of factor combinations
factors_combination = []

# recursive function
def compute_factors(current_no, n, product, single_list):
    global factors_combination

    # base case: if the product
    # exceeds our given number;
    # OR
    # current_no exceeds half the given n
    if ((current_no > int(n / 2)) or (product > n)):
        return

    # if current list of factors
    # is contributing to n
    if (product == n):
        # storing the list
        factors_combination.append(single_list)

        # printing all possible factors stored in
        # factors_combination
        for i in range(len(factors_combination)):
            for j in range(len(factors_combination[i])):
                print(factors_combination[i][j], end=" ")
        print()
        factors_combination = []
        # into factors_combination
        return

    # including current_no in our list
    single_list.append(current_no)

    # trying to get required
    # n with including current
    # current_no
    compute_factors(current_no, n, product * current_no, single_list)

    # excluding current_no from our list
    single_list.pop()

    # trying to get required n
    # without including current
    # current_no
    compute_factors(current_no + 1, n, product, single_list)

n = 16

# vector to store single list of factors
# eg. 2,2,2,2 is one of the list for n=16
single_list = []

# compute_factors ( starting_no, given_n,
# our_current_product, vector )
compute_factors(2, n, 1, single_list)

# This code is contributed by ukasp.

C#

// C# program to print all factors combination
using System;
using System.Collections.Generic;
class GFG {

    // vector of vector for storing
    // list of factor combinations
    static List<List<int>> factors_combination = new List<List<int>>();

    // recursive function
    static void compute_factors(int current_no, int n, int product, List<int> single_list)
    {

        // base case: if the product
        // exceeds our given number;
        // OR
        // current_no exceeds half the given n
        if (current_no > (n / 2) || product > n)
            return;

        // if current list of factors
        // is contributing to n
        if (product == n) {

            // storing the list
            factors_combination.Add(single_list);
            // printing all possible factors stored in
            // factors_combination
            for(int i = 0; i < factors_combination.Count; i++)
            {
                for(int j = 0; j < factors_combination[i].Count; j++)
                Console.Write(factors_combination[i][j] + " ");
            }
            Console.WriteLine();
            factors_combination = new List<List<int>>();
            // into factors_combination
            return;
        }

        // including current_no in our list
        single_list.Add(current_no);

        // trying to get required
        // n with including current
        // current_no
        compute_factors(current_no, n, product * current_no, single_list);

        // excluding current_no from our list
        single_list.RemoveAt(single_list.Count - 1);

        // trying to get required n
        // without including current
        // current_no
        compute_factors(current_no + 1, n, product, single_list);
    }

  static void Main() {
    int n = 16;

    // vector to store single list of factors
    // eg. 2,2,2,2 is one of the list for n=16
    List<int> single_list = new List<int>();

    // compute_factors ( starting_no, given_n,
    // our_current_product, vector )
    compute_factors(2, n, 1, single_list);
  }
}

// This code is contributed by divyesh072019.

Javascript

<script>
    // Javascript program to print all factors combination

    // vector of vector for storing
    // list of factor combinations
    let factors_combination = [];

    // recursive function
    function compute_factors(current_no, n, product, single_list)
    {
        // base case: if the product
        // exceeds our given number;
        // OR
        // current_no exceeds half the given n
        if ((current_no > parseInt(n / 2, 10)) || (product > n))
            return;

        // if current list of factors
        // is contributing to n
        if (product == n)
        {
            // storing the list
            factors_combination.push(single_list);

            // printing all possible factors stored in
            // factors_combination
            for(let i = 0; i < factors_combination.length; i++)
            {
                for(let j = 0; j < factors_combination[i].length; j++)
                {
                    document.write(factors_combination[i][j] + " ");
                }
            }
            document.write("</br>");
            factors_combination = [];
            // into factors_combination
            return;
        }

        // including current_no in our list
        single_list.push(current_no);

        // trying to get required
        // n with including current
        // current_no
        compute_factors(current_no, n, product * current_no, single_list);

        // excluding current_no from our list
        single_list.pop();

        // trying to get required n
        // without including current
        // current_no
        compute_factors(current_no + 1, n, product, single_list);
    }

    let n = 16;

    // vector to store single list of factors
    // eg. 2,2,2,2 is one of the list for n=16
    let single_list = [];

    // compute_factors ( starting_no, given_n,
    // our_current_product, vector )
    compute_factors(2, n, 1, single_list);

// This code is contributed by suresh07.
</script>

Output

2 2 2 2 
2 2 4 
2 8 
4 4 

Time Complexity: O(n²) , n is the size of vector
Auxiliary Space: O(n²), n is the size of vector

Please suggest if someone has a better solution which is more efficient in terms of space and time.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above