Find four factors of N with maximum product and sum equal to N - Set-2 in C++

Concept

With respect of a given integer N, our task is to determine all factors of N and print the product of four factors of N so that −

• The sum of the four factors is equal to N.
• The product of the four factors is largest.

It has been seen that if it is impossible to determine 4 such factors then print “Not possible”. It should be noted thatall the four factors can be equal to each other to maximize the product.

Input

N = 60

Output

All the factors are -> 1 2 3 4 5 6 10 12 15 20 30 60
Product is -> 50625

Select the factor 15 four times,

Therefore, 15+15+15+15 = 60 and product is largest.

Method

Here a method which takes a complexity of O(P^3), where P is the number of factors of N has been explained.

So an efficient method of time complexity O(N^2) can be obtained with the help of following steps.

• We store all the factors of given number in a container.
• Now we iterate for all pairs and store their sum in a different container.
• We have to mark the index (element1 + element2) with pair(element1, element2) to obtain the elements by which the sum was obtained.
• Again we iterate for all the pair_sums, and verify if n-pair_sum exists in the same container, as a result both the pairs form the quadruple.
• Implement the pair hash array to obtain the elements by which the pair was formed.
• Finally, store the largest of all such quadruples, and print it at the end.

Example

 Live Demo

// C++ program to find four factors of N
// with maximum product and sum equal to N
#include <bits/stdc++.h>
using namespace std;
// Shows function to find factors
// and to print those four factors
void findfactors1(int q){
   vector<int> vec1;
   // Now inserting all the factors in a vector s
   for (int i = 1; i * i <= q; i++) {
      if (q % i == 0) {
         vec1.push_back(i);
         vec1.push_back(q / i);
      }
   }
   // Used to sort the vector
   sort(vec1.begin(), vec1.end());
   // Used to print all the factors
   cout << "All the factors are -> ";
   for (int i = 0; i < vec1.size(); i++)
      cout << vec1[i] << " ";
      cout << endl;
      // So any elements is divisible by 1
      int maxProduct1 = 1;
      bool flag1 = 1;
      // implementing three loop we'll find
      // the three largest factors
      for (int i = 0; i < vec1.size(); i++) {
         for (int j = i; j < vec1.size(); j++) {
            for (int k = j; k < vec1.size(); k++) {
               // Now storing the fourth factor in y
               int y = q - vec1[i] - vec1[j] - vec1[k];
               // It has been seen that if the fouth factor become negative
               // then break
            if (y <= 0)
               break;
            // So we will replace more optimum number
            // than the previous one
            if (q % y == 0) {
               flag1 = 0;
               maxProduct1 = max(vec1[i] * vec1[j] * vec1[k] *y,maxProduct1);
            }
         }
      }
   }
   // Used to print the product if the numbers exist
   if (flag1 == 0)
      cout << "Product is -> " << maxProduct1 << endl;
   else
      cout << "Not possible" << endl;
}
// Driver code
int main(){
   int q;
   q = 60;
   findfactors1(q);
   return 0;
}

Output

All the factors are -> 1 2 3 4 5 6 10 12 15 20 30 60
Product is -> 50625