split a value into values in max, min range - python-3.x

Question

I want to find an efficient algorithm to divide an integer number to some value in a max, min range. There should be as less values as possible.
For example:
max = 7, min = 3
then
8 = 4 + 4
9 = 4 + 5
16 = 5 + 5 + 6 (not 4 + 4 + 4 + 4)
EDIT
To make it more clear, let take an example. Assume that you have a bunch of apples and you want to pack them into baskets. Each basket can contain 3 to 7 apples, and you want the number of baskets to be used is as small as possible.
** I mentioned that the value should be evenly divided, but that's not so important. I am more concerned about less number of baskets.

Answer 1

This struck me as a fun problem so I had a go at hacking out a quick solution. I think this might be an interesting starting point, it'll either give you a valid solution with as few numbers as possible, or with numbers as similar to each other as possible, all within the bounds of the range defined by the min_bound and max_bound
number = int(input("Number: "))
min_bound = 3
max_bound = 7
def improve(solution):
solution = list(reversed(solution))
for i, num in enumerate(solution):
if i >= 2:
average = sum(solution[:i]) / i
if average.is_integer():
for x in range(i):
solution[x] = int(average)
break
return solution
def find_numbers(number, division, common_number):
extra_number = number - common_number * division
numbers_in_solution = [common_number] * division
if extra_number < min_bound and \
extra_number + common_number <= max_bound:
numbers_in_solution[-1] += extra_number
elif extra_number < min_bound or extra_number > max_bound:
return None
else:
numbers_in_solution.append(extra_number)
solution = improve(numbers_in_solution)
return solution
def tst(number):
try:
solution = None
for division in range(number//max_bound, number//min_bound + 1): # Reverse the order of this for numbers as close in value to each other as possible.
if round (number / division) in range(min_bound, max_bound + 1):
solution = find_numbers(number, division, round(number / division))
elif (number // division) in range(min_bound, max_bound + 1): # Rarely required but catches edge cases
solution = find_numbers(number, division, number // division)
if solution:
print(sum(solution), solution)
break
except ZeroDivisionError:
print("Solution is 1, your input is less than the max_bound")
tst(number)
for x in range(1,100):
tst(x)
This code is just to demonstrate an idea, I'm sure it could be tweaked for better performance.