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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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    \$\begingroup\$ Great first question! \$\endgroup\$ Commented Nov 23, 2020 at 13:06
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    \$\begingroup\$ Can you provide a link to the original question? \$\endgroup\$ Commented Nov 23, 2020 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\$ Commented Nov 24, 2020 at 0:58

2 Answers 2

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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\$ Commented Nov 24, 2020 at 1:00
  • \$\begingroup\$ A good name? For which variable? \$\endgroup\$ Commented Nov 24, 2020 at 1:01
  • \$\begingroup\$ Storage of the Fibonacci numbers. \$\endgroup\$ Commented Nov 24, 2020 at 1:09
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    \$\begingroup\$ Oh, you mean x0 and x1. prev and current maybe? \$\endgroup\$ Commented Nov 24, 2020 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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    \$\begingroup\$ Ish. Overall this is a good suggestion, but framing this as a list doesn't make sense. Just have two variables. \$\endgroup\$ Commented Nov 23, 2020 at 17:03
  • \$\begingroup\$ In other words, keep the OP's x0, x1, possibly with better names. \$\endgroup\$ Commented Nov 23, 2020 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\$ Commented Nov 24, 2020 at 1:06

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