Google foobar level 2 challenge

Question

I am a beginner in Python-3. I was solving a level 2 challenge in Google Foobar. Given an initial amount of items (LAMBs) I have to distribute it among individuals, who are ranked according to seniority, while adhering to the following rules:

1. The junior most individual gets 1 Lamb
2. An immediately senior individual can't get more than twice their juniors lambs, otherwise the junior will rebel
3. A senior individual can't get less lambs than the sum of their 2 previous juniors, otherwise the senior will rebel
4. In case lambs are leftover and adding a senior does not violate any previous rules, they must be added.

The goal is to find the difference between how many individuals can be paid if we pay the minimum amount to each (stingy) and the maximum amount to each (generous). I noticed that stingy condition resembles a Fibonacci sequence and the generous condition is sum of powers of 2. My code is as follows:

from math import log2

def solution(total_lambs)
    return (stingy(total_lambs) - generous(total_lambs))

def stingy(total_lambs)
    if (total_lambs == 1): # from condition 1 at least 1 individual gets paid
        return 1
    elif (total_lambs == 2): # from condition 2 and 4 the most each can is 1
        return 2
    else:
        ind, total, x0, x1 = 2, 2, 1, 1 # no. individuals, sum of Fib seq, intial Fib nos.
        while (total <= total_lambs):
            x0, x1 = x1, x0 + x1 # advancing the Fib seq
            total = total + x1 # total is the sum of lambs paid
            if (total > total_lambs): 
                return ind
            else:
                ind = ind + 1

def generous(total_lambs):
    if (total_lambs == 1):
        return 1
    if (total_lambs == 2):
        return 2
    else:
        ind = int(log2(capital + 1)) # typecasting to int removes decimals
        if (total_lambs - 2**n + 1 >= 2**(n - 1) + 2**(n - 2)): # condition 2, 3 and 4 
            return ind + 1
        else:
            return ind

Please suggest improvements wherever possible.

Answer 1

Early-returns

else-after-return is redundant, so your stingy is equivalent to

    if (total_lambs == 1): # from condition 1 at least 1 individual gets paid
        return 1
    if (total_lambs == 2): # from condition 2 and 4 the most each can is 1
        return 2

    ind, total, x0, x1 = 2, 2, 1, 1 # no. individuals, sum of Fib seq, intial Fib nos.
    while (total <= total_lambs):
        x0, x1 = x1, x0 + x1 # advancing the Fib seq
        total = total + x1 # total is the sum of lambs paid
        if (total > total_lambs): 
            return ind
        ind = ind + 1

Return parens

You have a mix of this style:

return 1

and this style:

return (stingy(total_lambs) - generous(total_lambs))

the latter not needing outer parens. Probably best to go with the non-parenthetical style. The same is true of

while (total <= total_lambs):

In-place addition

total = total + x1
ind = ind + 1

can be

total += x1
ind += 1

Answer 2

apart from the improvements mentioned above, for the stingy function you can also reduce the amount of code by using a list with the last 2 fibonacci members, and integrating the initial conditions into the logic

def stingy(total_lambs):
  paid = 0
  previous = [1, 0]
  while total_lambs > 0:
    paid += 1
    previous = [sum(previous), previous[0]]
    total_lambs -= previous[0]
  return paid