How can I make my code for calculating what is the minimum possible ways to divide each item by 2 faster?

Question

Giving a list of numbers A , the goal is to find the minimum possible number of divisions you can do on each item to get an x in which x <= sum(A)/2 .

For example,

• if A = [10,10] you can divide the A[0] by 2 and you can divide A[1] by 2 as well to get half of the sum of the list. So, the minimum possible number of divisions equals to 2 ,
• if A = [200, 25,25,25] you can divide A[0]/2^3 so x =[25,25,25,25] which sum to 100<= 275 so minimum possible number of divisions equals to 3 .

I tried the following way and it works just fine, but when I input huage lists like this get_f([200,25,25,25,490,99999, 2000, 43002]) I got timeout error.

import itertools

def get_f(A):
    
  A = sorted(A)[::-1]
  Goal = sum(A)/2

  filters = []
  x = itertools.product(range(len(A)), repeat=len(A))
  for i in x:
    i = list(i)
    C = A.copy()
    x = 0
    for Index in i:
      C[Index]/=2
      x+=1
      if sum(C) <= Goal and x not in filters:
        filters.append(x)
  return sorted(filters)[0]

Answer 1

The currently accepted solution is incorrect- it will, for instance, never return a number larger than the length of the input list, regardless of how large any of those numbers are.

The optimal way to approach this problem is with a priority queue- allowing you to avoid recomputing the max() (as done in another answer) over and over.

Code:

from heapq import heapify, heappop, heappush
from itertools import count
from math import isclose

def solve(nums):
    heap = [-x for x in nums]
    heapify(heap)

    total = sum(nums)
    target = total / 2

    for divs in count():
        if total < target or isclose(total, target):
            return divs
        
        delta = heappop(heap) / 2
        total += delta
        heappush(heap, delta)

Demo:

>>> print(solve([200, 25, 25, 25]))
2
>>> print(solve([200, 25, 25, 25, 0]))
2
>>> print(solve([100, 25, 100, 25]))
3

Note : the original example in the problem statement is wrong- you can divide 200 by 2 twice, and the sum [50, 25, 25, 25] = 125 is less than the original sum 275 , making the answer 2 , not 3 .

Answer 2

Here is my version of the code. It works just fine on the tests. Check it out.

def get_f(A):
    """
    A: list of numbers
    return: int
    """
    x = sum(A) // 2
    if x == 0:
        return 0
    count = 0
    for i in A:
        if i <= x:
            count += 1
    return count

print(get_f([10,10]))
print(get_f([200, 25,25,25]))
print(get_f([10,10,10]))
print(get_f([10,10,10,10]))
print(get_f([200,25,25,25,490,99999, 2000, 43002]))

Hope this is what you were looking for

Answer 3

def get_f(xs):
    goal = sum(xs) / 2
    f = 0
    while sum(xs) > goal:
        i = xs.index(max(xs))
        xs[i] /= 2
        f += 1
    return f

This should work in case it's okay to mutate the input list.