Andrzej was given a task:
There are n jars with pills. In every jar there is a different type of pill and the amount of pills in each jar is infinite. One type of pill makes a person glow about 30 minutes after taking and none of the other types has any effect.
His job is to determine, in which jar are the pills that make a person glow.
But there is one catch, he only has 35 minutes to do so.(so he can't take a pill, wait for the results and then take another one, because he wouldn't be able to see the results)
Fortunetely, he can take any number of friends he needs with him.
On completing the task Andrzej receives one million dollars. You know that Andrzej is very honest, so he will split the money equally with his friends.
Your job is to determine how many friends does Andrzej need to complete the task.(He also wants to make the highest amount of money.)
For example for n = 2
The answer is 0 because he doesn't need any friends, he just needs to take a pill from the first jar and wait for the effects.
For another example for n = 4
The answer is 1 because having pills A B C D Andrzej can take pills A B and the friend can take pills B C

Difficulty: introductory

Starter code:

def friends(n):
	

Solutions:
def friends(n):
    return len(str(bin(n-1)))-3 if n >1 else 0

def friends(n):
    return (n-1 or 1).bit_length()-1

def friends(n):
    return 0 if n < 2 else (n - 1).bit_length() - 1

from math import log2

def friends(n):
    return int(log2(max(2, n) - 1))

from math import log2

def friends(n):
    return int(log2(n-1)) if n>1 else 0

def relu(n):
    return 0 if n < 0 else n

def friends(n):
    return relu(~-(~-n).bit_length())

from math import log,ceil
def friends(n): return 0 if n < 2 else ceil(log(n,2)-1)

import math
def friends(n):
    print(n)
    if n <= 2:
        return 0
    elif math.log(n,2)==int(math.log(n,2)):
        return int(math.log(n,2)) -1
    else:
        return int(math.log(n,2))


import math

def friends(n):
    if n <= 1:
        return 0
    else:
        return math.ceil(math.log(n) / math.log(2))-1

def friends(n):
    x = 1
    while True:
        if n / (2**x) <= 1:
            break
        x += 1
    return x - 1