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

PythonServer Side ProgrammingProgramming

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

60

## Output

Maximum product 50625
Published on 25-Aug-2020 09:15:48