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

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:

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// 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;
}

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Output:

The maximum product is 12500


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Striver(underscore)79 at Codechef and codeforces D

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Improved By : Akanksha_Rai