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Find four factors of N with maximum product and sum equal to N | Set-2

  • Last Updated : 27 May, 2021

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

  • Sum of the four factors is equal to N.
  • 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.
Examples
 

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

 

An approach which takes a complexity of O(M^3), where M is the number of factors of N has been discussed in the previous post. 
An efficient approach of time complexity O(N^2) can be obtained by following the below steps. 
 

  • Store all the factors of given number in a container.
  • Iterate for all pairs and store their sum in a different container.
  • Mark the index (element1 + element2) with pair(element1, element2 to get the elements by which the sum was obtained.
  • Iterate for all the pair_sums, and check if n-pair_sum exists in the same container, then both the pairs form the quadruple.
  • Use the pair hash array to get the elements by which the pair was formed.
  • Store the maximum of all such quadruples, and print it at the end.

Below is the implementation of above approach: 
 

C++




// C++ program to find four factors of N
// with maximum product and sum equal to N
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to find factors
// and to print those four factors
void findfactors(int n)
{
    unordered_map<int, int> mpp;
 
    vector<int> v, v1;
 
    // push all the factors in the container
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            v.push_back(i);
            if (i != (n / i) && i != 1)
                v.push_back(n / i);
        }
    }
 
    // number of factors
    int s = v.size();
 
    // Initial maximum
    int maxi = -1;
 
    // hash-array to mark the
    // pairs
    pair<int, int> mp1[n + 5];
 
    for (int i = 0; i < s; i++) {
 
        // form all the pair sums
        for (int j = i; j < s; j++) {
 
            // if the pair sum is less than n
            if (v[i] + v[j] < n) {
 
                // push in another container
                v1.push_back(v[i] + v[j]);
 
                // mark the sum with the elements
                // formed
                mp1[v[i] + v[j]] = { v[i], v[j] };
 
                // mark in the map that v[i]+v[j]
                // is present
                mpp[v[i] + v[j]] = 1;
            }
        }
    }
 
    // new size of all the pair sums
    s = v1.size();
 
    // iterate for all pair sum
    for (int i = 0; i < s; i++) {
 
        // the required part
        int el = n - (v1[i]);
 
        // if the required part is also
        // present in pair sum
        if (mpp[el] == 1) {
 
            // find the elements with
            // which the first pair is formed
            int a = mp1[v1[i]].first;
            int b = mp1[v1[i]].second;
 
            // find the elements with
            // which the second pair is formed
            int c = mp1[n - v1[i]].first;
            int d = mp1[n - v1[i]].second;
 
            // check for previous maximum
            maxi = max(a * b * c * d, maxi);
        }
    }
 
    if (maxi == -1)
        cout << "Not Possible\n";
    else {
        cout << "The maximum product is " << maxi << endl;
    }
}
 
// Driver code
int main()
{
    int n = 50;
 
    findfactors(n);
 
    return 0;
}

Java




// Java program to find four factors of N
// with maximum product and sum equal to N
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
     
// Function to find factors
// and to print those four factors
static void findfactors(int n)
{
    HashMap<Integer,
            Integer> mpp = new HashMap<>();
 
    Vector<Integer> v = new Vector<Integer>(),
                   v1 = new Vector<Integer>();
 
    // push all the factors in the container
    for (int i = 1; i <= (int)Math.sqrt(n); i++)
    {
        if (n % i == 0)
        {
            v.add(i);
            if (i != (n / i) && i != 1)
                v.add(n / i);
        }
    }
 
    // number of factors
    int s = v.size();
 
    // Initial maximum
    int maxi = -1;
 
    // hash-array to mark the
    // pairs
    int mp1_first[] = new int[n + 5],
        mp1_second[] = new int[n + 5];
 
    for (int i = 0; i < s; i++)
    {
 
        // form all the pair sums
        for (int j = i; j < s; j++)
        {
 
            // if the pair sum is less than n
            if (v.get(i) + v.get(j) < n)
            {
 
                // push in another container
                v1.add(v.get(i) + v.get(j));
 
                // mark the sum with the elements
                // formed
                mp1_first[v.get(i) +
                          v.get(j)] = v.get(i);
                mp1_second[v.get(i) +
                           v.get(j)] = v.get(j);
 
                // mark in the map that
                // v.get(i)+v.get(j) is present
                mpp.put(v.get(i) + v.get(j), 1);
            }
        }
    }
 
    // new size of all the pair sums
    s = v1.size();
 
    // iterate for all pair sum
    for (int i = 0; i < s; i++)
    {
 
        // the required part
        int el = n - (v1.get(i));
 
        // if the required part is also
        // present in pair sum
        if (mpp.get(el) != null)
        {
 
            // find the elements with
            // which the first pair is formed
            int a = mp1_first[v1.get(i)];
            int b = mp1_second[v1.get(i)];
 
            // find the elements with
            // which the second pair is formed
            int c = mp1_first[n - v1.get(i)];
            int d = mp1_second[n - v1.get(i)];
 
            // check for previous maximum
            maxi = Math.max(a * b * c * d, maxi);
        }
    }
 
    if (maxi == -1)
        System.out.println("Not Possible");
    else
    {
        System.out.println("The maximum product" +
                                   " is " + maxi);
    }
}
 
// Driver code
public static void main(String args[])
{
    int n = 50;
 
    findfactors(n);
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 program to find four factors of N
# with maximum product and sum equal to N
from math import sqrt, ceil, floor
 
# Function to find factors
# and to prthose four factors
def findfactors(n):
    mpp = dict()
 
    v = []
    v1 = []
 
    # push all the factors in the container
    for i in range(1,ceil(sqrt(n)) + 1):
        if (n % i == 0):
            v.append(i)
            if (i != (n // i) and i != 1):
                v.append(n // i)
 
    # number of factors
    s = len(v)
 
    # Initial maximum
    maxi = -1
 
    # hash-array to mark the
    # pairs
    mp1 = [0]*(n + 5)
 
    for i in range(s):
 
        # form all the pair sums
        for j in range(i, s):
 
            # if the pair sum is less than n
            if (v[i] + v[j] < n):
 
                # push in another container
                v1.append(v[i] + v[j])
 
                # mark the sum with the elements
                # formed
                mp1[v[i] + v[j]] =[v[i], v[j]]
 
                # mark in the map that v[i]+v[j]
                # is present
                mpp[v[i] + v[j]] = 1
 
 
    # new size of all the pair sums
    s = len(v1)
 
    # iterate for all pair sum
    for i in range(s):
 
        # the required part
        el = n - (v1[i])
 
        # if the required part is also
        # present in pair sum
        if (el in mpp):
 
            # find the elements with
            # which the first pair is formed
            a = mp1[v1[i]][0]
            b = mp1[v1[i]][1]
 
            # find the elements with
            # which the second pair is formed
            c = mp1[n - v1[i]][0]
            d = mp1[n - v1[i]][1]
 
            # check for previous maximum
            maxi = max(a * b * c * d, maxi)
 
    if (maxi == -1):
        print("Not Possible")
    else :
        print("The maximum product is ", maxi)
 
# Driver code
n = 50
 
findfactors(n)
 
# This code is contributed by mohit kumar 29

C#




// C# program to find four factors of N
// with maximum product and sum equal to N
using System;
using System.Collections.Generic;
 
class GFG
{
     
// Function to find factors
// and to print those four factors
static void findfactors(int n)
{
    Dictionary<int,
            int> mpp = new Dictionary<int,int>();
 
    List<int> v = new List<int>(),
                v1 = new List<int>();
 
    // push all the factors in the container
    for (int i = 1; i <= (int)Math.Sqrt(n); i++)
    {
        if (n % i == 0)
        {
            v.Add(i);
            if (i != (n / i) && i != 1)
                v.Add(n / i);
        }
    }
 
    // number of factors
    int s = v.Count;
 
    // Initial maximum
    int maxi = -1;
 
    // hash-array to mark the
    // pairs
    int []mp1_first = new int[n + 5];
    int []mp1_second = new int[n + 5];
 
    for (int i = 0; i < s; i++)
    {
 
        // form all the pair sums
        for (int j = i; j < s; j++)
        {
 
            // if the pair sum is less than n
            if (v[i] + v[j] < n)
            {
 
                // push in another container
                v1.Add(v[i] + v[j]);
 
                // mark the sum with the elements
                // formed
                mp1_first[v[i] +
                        v[j]] = v[i];
                mp1_second[v[i] +
                        v[j]] = v[j];
 
                // mark in the map that
                // v[i]+v[j] is present
                mpp.Add(v[i] + v[j], 1);
            }
        }
    }
 
    // new size of all the pair sums
    s = v1.Count;
 
    // iterate for all pair sum
    for (int i = 0; i < s; i++)
    {
 
        // the required part
        int el = n - (v1[i]);
 
        // if the required part is also
        // present in pair sum
        if (mpp.ContainsKey(el))
        {
 
            // find the elements with
            // which the first pair is formed
            int a = mp1_first[v1[i]];
            int b = mp1_second[v1[i]];
 
            // find the elements with
            // which the second pair is formed
            int c = mp1_first[n - v1[i]];
            int d = mp1_second[n - v1[i]];
 
            // check for previous maximum
            maxi = Math.Max(a * b * c * d, maxi);
        }
    }
 
    if (maxi == -1)
        Console.WriteLine("Not Possible");
    else
    {
        Console.WriteLine("The maximum product" +
                                " is " + maxi);
    }
}
 
// Driver code
public static void Main(String []args)
{
    int n = 50;
 
    findfactors(n);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
// Javascript program to find four factors of N
// with maximum product and sum equal to N
 
// Function to find factors
// and to print those four factors
function findfactors(n)
{
    let mpp = new Map();
   
    let v = [];
    let v1 = [];
   
    // push all the factors in the container
    for (let i = 1; i <= Math.floor(Math.sqrt(n)); i++)
    {
        if (n % i == 0)
        {
            v.push(i);
            if (i != Math.floor(n / i) && i != 1)
                v.push(Math.floor(n / i));
        }
    }
   
    // number of factors
    let s = v.length;
   
    // Initial maximum
    let maxi = -1;
   
    // hash-array to mark the
    // pairs
    let mp1_first = new Array(n + 5),
        mp1_second = new Array(n + 5);
       
    for (let i = 0; i < s; i++)
    {
   
        // form all the pair sums
        for (let j = i; j < s; j++)
        {
   
            // if the pair sum is less than n
            if (v[i] + v[j] < n)
            {
   
                // push in another container
                v1.push(v[i] + v[j]);
   
                // mark the sum with the elements
                // formed
                mp1_first[v[i] +
                          v[j]] = v[i];
                mp1_second[v[i] +
                           v[j]] = v[j];
   
                // mark in the map that
                // v.get(i)+v.get(j) is present
                mpp.set(v[i] + v[j], 1);
            }
        }
    }
   
    // new size of all the pair sums
    s = v1.length;
   
    // iterate for all pair sum
    for (let i = 0; i < s; i++)
    {
   
        // the required part
        let el = n - (v1[i]);
   
        // if the required part is also
        // present in pair sum
        if (mpp.get(el) != null)
        {
   
            // find the elements with
            // which the first pair is formed
            let a = mp1_first[v1[i]];
           let b = mp1_second[v1[i]];
   
            // find the elements with
            // which the second pair is formed
            let c = mp1_first[n - v1[i]];
            let d = mp1_second[n - v1[i]];
   
            // check for previous maximum
            maxi = Math.max(a * b * c * d, maxi);
        }
    }
   
    if (maxi == -1)
        document.write("Not Possible<br>");
    else
    {
        document.write("The maximum product" +
                                   " is " + maxi);
    }
}
 
// Driver code
let n = 50;
findfactors(n);
     
 
// This code is contributed by patel2127
</script>
Output: 
The maximum product is 12500

 

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