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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.

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  • 2
    \$\begingroup\$ Great first question! \$\endgroup\$
    – Reinderien
    Nov 23 '20 at 13:06
  • 3
    \$\begingroup\$ Can you provide a link to the original question? \$\endgroup\$
    – Reinderien
    Nov 23 '20 at 13:07
  • \$\begingroup\$ @reinderien I unfortunately submitted my response before i saved the original question so i can't copy it verbatim. However, it is available online if you search foobar Lovely lucky LAMBs. \$\endgroup\$
    – Roni Saiba
    Nov 24 '20 at 0:58
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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
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  • \$\begingroup\$ Thank you. As you mentioned in another comment, what can be a good name for storing Fibonnacci numbers? \$\endgroup\$
    – Roni Saiba
    Nov 24 '20 at 1:00
  • \$\begingroup\$ A good name? For which variable? \$\endgroup\$
    – Reinderien
    Nov 24 '20 at 1:01
  • \$\begingroup\$ Storage of the Fibonacci numbers. \$\endgroup\$
    – Roni Saiba
    Nov 24 '20 at 1:09
  • 1
    \$\begingroup\$ Oh, you mean x0 and x1. prev and current maybe? \$\endgroup\$
    – Reinderien
    Nov 24 '20 at 1:25
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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
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  • 1
    \$\begingroup\$ Ish. Overall this is a good suggestion, but framing this as a list doesn't make sense. Just have two variables. \$\endgroup\$
    – Reinderien
    Nov 23 '20 at 17:03
  • \$\begingroup\$ In other words, keep the OP's x0, x1, possibly with better names. \$\endgroup\$
    – Reinderien
    Nov 23 '20 at 17:05
  • \$\begingroup\$ Thanks, i will keep this structure in mind for future coding. I framed my function in this way as the rule 3 in original question had an explanation that it is not applicable if only upto 2 individuals get paid. So I did not think of this elegant way. \$\endgroup\$
    – Roni Saiba
    Nov 24 '20 at 1:06

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