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

Python Server Side Programming Programming

Suppose we have a number N; we have to find factors of N and return only the product of four factors of N such that −

The sum of the four factors is same as N.
The product of the four factors is maximum.

All four factors can be equal to each other to maximize the product.

So, if the input is like N = 60, then the output will be 50625 as all the factors are → 1 2 3 4 5 6 10 12 15 20 30 60 and their product is 50625, as we 15 has been selected four times to make product largest.

To solve this, we will follow these steps −

my_map := a new map
v := a new list, v1 := a new list
for i in range 1 to the ceiling of square root of (n) + 1, do
- if n mod i is same as 0, then
  - insert i at the end of v
  - if i is not same as integer part of (n / i) and i is not same as 1, then
    - insert integer part of (n / i) at the end of v
s := size of v
maximum := -1
map1 := an array of size (n + 5) fill this with 0
for i in range 0 to s, do
- for j in range i to s, do
  - if v[i] + v[j] < n , then
    - insert v[i] + v[j] at the end of v1
    - map1[v[i] + v[j]] := [v[i], v[j]]
    - my_map[v[i] + v[j]] := 1
s := size of v1
for i in range 0 to s, do
- element := n - (v1[i])
- if element in my_map, then
  - a := map1[v1[i], 0]
  - b := map1[v1[i], 1]
  - c := map1[n - v1[i], 0]
  - d := map1[n - v1[i], 1]
  - maximum := maximum of a * b * c * d, maximum
if maximum is same as -1, then
- display "Not Possible"
otherwise,
- display maximum

Example

Let us see the following implementation to get better understanding −

Live Demo

from math import sqrt, ceil, floor
def get_product(n):
   my_map = dict()
   v = []
   v1 = []
   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)
   s = len(v)
   maximum = -1
   map1 = [0]*(n + 5)
   for i in range(s):
      for j in range(i, s):
         if (v[i] + v[j] < n):
            v1.append(v[i] + v[j])
            map1[v[i] + v[j]] =[v[i], v[j]]
            my_map[v[i] + v[j]] = 1
   s = len(v1)
   for i in range(s):
      element = n - (v1[i])
      if (element in my_map):
         a = map1[v1[i]][0]
         b = map1[v1[i]][1]
         c = map1[n - v1[i]][0]
         d = map1[n - v1[i]][1]
         maximum = max(a * b * c * d, maximum)
   if (maximum == -1):
      print("Not Possible")
   else :
      print("Maximum product", maximum)
n = 60
get_product(n)

Input

Output

Maximum product 50625

Arnab Chakraborty

Published on 25-Aug-2020 09:15:48

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