Find four factors of N with maximum product and sum equal to N | Set 3

Given an integer N. The task is to find all factors of N and print the product of four factors of N such that:

1. Sum of the four factors is equal to N.
2. The product of the four factors is maximum.

If it is not possible to find 4 such factors then print “Not possible”.
Note: All the four factors can be equal to each other to maximize the product and there can be a large number of queries.
Examples: 

Input: 24
Output: Product -> 1296
All factors are -> 1 2 3 4 6 8 12 24 
Choose the factor 6 four times,
Therefore, 6+6+6+6 = 24 and product is maximum.

Input: 100
Output: Product -> 390625
All the factors are -> 1 2 4 5 10 10 20 25 50 100 
Choose the factor 25 four times.

The idea is to find factors of all numbers from 1 to N ( which is the maximum value of n ). 

• An answer will be Not possible if the given is prime.
• And if the given n is divisible by 4 then answer will be pow(q, 4) where q is a quotient when n is divided by 4.
• If it is possible to find the answer then, the answer must include third last factor two times. And run a nested loop for other two factors.

Below is the implementation of the above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find primes
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Function to find factors
void factors(int N, vector<int>& v[])
{
    for (int i = 2; i < N; i++) {

        // run a loop upto square root of that number
        for (int j = 1; j * j <= i; j++) {
            if (i % j == 0) {

                // if the n is perfect square
                if (i / j == j)
                    v[i].push_back(j);

                // otherwise push it's two divisors
                else {
                    v[i].push_back(j);
                    v[i].push_back(i / j);
                }
            }
        }

        // sort the divisors
        sort(v[i].begin(), v[i].end());
    }
}

// Function to find max product
int product(int n)
{
    // To store factors of 'n'
    vector<int> v[n + 100];

    // find factors
    factors(n + 100, v);

    // if it is divisible by 4.
    if (n % 4 == 0) {
        int x = n / 4;
        x *= x;
        return x * x;
    }

    else {

        // if it is prime
        if (isPrime[n])
            return -1;

        // otherwise answer will be possible
        else {
            int ans = -1;
            if (v[n].size() > 2) {

                // include last third factor
                int fac = v[n][v[n].size() - 3];

                // nested loop to find other two factors
                for (int i = v[n].size() - 1; i >= 0; i--) {
                    for (int j = v[n].size() - 1; j >= 0; j--) {
                        if ((fac * 2) + (v[n][j] + v[n][i]) == n)
                            ans = max(ans, fac * fac * v[n][j] * v[n][i]);
                    }
                }

                return ans;
            }
        }
    }
}

// Driver code
int main()
{

    int n = 24;

    // function call
    cout << product(n);

    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{

// Function to find primes
static boolean isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

static Vector<Vector<Integer> > v = new Vector<Vector<Integer> >();

// Function to find factors
static void factors(int N )
{
    for (int i = 2; i < N; i++)
    {

        // run a loop upto square root of that number
        for (int j = 1; j * j <= i; j++)
        {
            if (i % j == 0)
            {

                // if the n is perfect square
                if (i / j == j)
                    v.get(i).add(j);

                // otherwise push it's two divisors
                else
                {
                    v.get(i).add(j);
                    v.get(i).add(i / j);
                }
            }
        }

        // sort the divisors
        Collections.sort(v.get(i));
    }
}

// Function to find max product
static int product(int n)
{
    // To store factors of 'n'
    v.clear();
    for(int i = 0; i < n + 100; i++)
        v.add(new Vector<Integer>());

    // find factors
    factors(n + 100);

    // if it is divisible by 4.
    if (n % 4 == 0)
    {
        int x = n / 4;
        x *= x;
        return x * x;
    }

    else
    {

        // if it is prime
        if (isPrime(n))
            return -1;

        // otherwise answer will be possible
        else
        {
            int ans = -1;
            if (v.get(n).size() > 2)
            {

                // include last third factor
                int fac = v.get(n).get(v.get(n).size() - 3);

                // nested loop to find other two factors
                for (int i = v.get(n).size() - 1; i >= 0; i--)
                {
                    for (int j = v.get(n).size() - 1; j >= 0; j--)
                    {
                        if ((fac * 2) + (v.get(n).get(j) +
                                         v.get(n).get(i)) == n)
                            ans = Math.max(ans, fac * fac *
                                          v.get(n).get(j) *
                                          v.get(n).get(i));
                    }
                }
                return ans;
            }
        }
    }
    return 0;
}

// Driver code
public static void main(String args[])
{
    int n = 24;

    // function call
    System.out.println( product(n));
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of above approach

from math import sqrt, ceil, floor

# Function to find primes
def isPrime(n):

    # Corner cases
    if (n <= 1):
        return False
    if (n <= 3):
        return True

    # This is checked so that we can skip
    # middle five numbers in below loop
    if (n % 2 == 0 or n % 3 == 0):
        return False

    for i in range(5, ceil(sqrt(n)), 6):
        if (n % i == 0 or n % (i + 2) == 0):
            return False

    return True

# Function to find factors
def factors(N, v):
    for i in range(2, N):

        # run a loop upto square root of that number
        for j in range(1,ceil(sqrt(i)) + 1):
            if (i % j == 0):

                # if the n is perfect square
                if (i // j == j):
                    v[i].append(j)

                # otherwise push it's two divisors
                else:
                    v[i].append(j)
                    v[i].append(i // j)

        # sort the divisors
        v = sorted(v)

# Function to find max product
def product(n):

    # To store factors of 'n'
    v = [[]] * (n + 100)

    # find factors
    factors(n + 100, v)

    # if it is divisible by 4.
    if (n % 4 == 0):
        x = n // 4
        x *= x
        return x * x

    else :

        # if it is prime
        if (isPrime[n]):
            return -1

        # otherwise answer will be possible
        else :
            ans = -1
            if (len(v[n]) > 2):

                # include last third factor
                fac = v[n][len(v[n]) - 3]

                # nested loop to find other two factors
                for i in range(len(v[n] - 1), -1, -1):
                    for j in range(len(v[n] - 1), -1, -1):
                        if ((fac * 2) + (v[n][j] + v[n][i]) == n):
                            ans = max(ans, fac * fac * v[n][j] * v[n][i])

                return ans

# Driver code
n = 24

# function call
print(product(n))

# This code is contributed by mohit kumar 29

C#

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

class GFG
{

// Function to find primes
static bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

static List<List<int> > v = new List<List<int> >();

// Function to find factors
static void factors(int N )
{
    for (int i = 2; i < N; i++)
    {

        // run a loop upto square root of that number
        for (int j = 1; j * j <= i; j++)
        {
            if (i % j == 0)
            {

                // if the n is perfect square
                if (i / j == j)
                    v[i].Add(j);

                // otherwise push it's two divisors
                else
                {
                    v[i].Add(j);
                    v[i].Add(i / j);
                }
            }
        }

        // sort the divisors
        v[i].Sort();
    }
}

// Function to find max product
static int product(int n)
{
    // To store factors of 'n'
    v.Clear();
    for(int i = 0; i < n + 100; i++)
        v.Add(new List<int>());

    // find factors
    factors(n + 100);

    // if it is divisible by 4.
    if (n % 4 == 0)
    {
        int x = n / 4;
        x *= x;
        return x * x;
    }

    else
    {

        // if it is prime
        if (isPrime(n))
            return -1;

        // otherwise answer will be possible
        else
        {
            int ans = -1;
            if (v[n].Count > 2)
            {

                // include last third factor
                int fac = v[n][v[n].Count - 3];

                // nested loop to find other two factors
                for (int i = v[n].Count - 1; i >= 0; i--)
                {
                    for (int j = v[n].Count - 1; j >= 0; j--)
                    {
                        if ((fac * 2) + (v[n][j] +
                                        v[n][i]) == n)
                            ans = Math.Max(ans, fac * fac *
                                        v[n][j] *
                                        v[n][i]);
                    }
                }
                return ans;
            }
        }
    }
    return 0;
}

// Driver code
public static void Main(String []args)
{
    int n = 24;

    // function call
    Console.WriteLine( product(n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript implementation of above approach

// Function to find primes
function isPrime(n) {
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (let i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Function to find factors
function factors(N, v) {
    for (let i = 2; i < N; i++) {

        // run a loop upto square root of that number
        for (let j = 1; j * j <= i; j++) {
            if (i % j == 0) {

                // if the n is perfect square
                if (i / j == j)
                    v[i].push(j);

                // otherwise push it's two divisors
                else {
                    v[i].push(j);
                    v[i].push(i / j);
                }
            }
        }

        // sort the divisors
        v.sort((a, b) => a - b);
    }
}

// Function to find max product
function product(n) {
    // To store factors of 'n'
    let v = new Array();

    for (let i = 0; i < n + 100; i++) {
        v.push(new Array())
    }
    // find factors
    factors(n + 100, v);

    // if it is divisible by 4.
    if (n % 4 == 0) {
        let x = n / 4;
        x *= x;
        return x * x;
    }

    else {

        // if it is prime
        if (isPrime[n])
            return -1;

        // otherwise answer will be possible
        else {
            let ans = -1;
            if (v[n].length > 2) {

                // include last third factor
                let fac = v[n][v[n].length - 3];

                // nested loop to find other two factors
                for (let i = v[n].length - 1; i >= 0; i--) {
                    for (let j = v[n].length - 1; j >= 0; j--) {
                        if ((fac * 2) + (v[n][j] + v[n][i]) == n)
                            ans = Math.max(ans, fac * fac * v[n][j] * v[n][i]);
                    }
                }

                return ans;
            }
        }
    }
}

// Driver code


let n = 24;

// function call
document.write(product(n));

// This code is contributed by _saurabh_jaiswal
</script>

Output:

1296

Time Complexity: O(n²)

Auxiliary Space: O(n²)