Problem

I am trying to solve this question from google’s foobar. However, my code exceeds the time limit, even though when I tested it I got correct answers. How can I make my code more efficient?

## Lovely Lucky LAMBs

Being a henchman isn’t all drudgery. Occasionally, when Commander

Lambda is feeling generous, she’ll hand out Lucky LAMBs (Lambda’s

All-purpose Money Bucks). Henchmen can use Lucky LAMBs to buy things

like a second pair of socks, a pillow for their bunks, or even a third

daily meal!However, actually passing out LAMBs isn’t easy. Each henchman squad

has a strict seniority ranking which must be respected – or else the

henchmen will revolt and you’ll all get demoted back to minions again!There are 4 key rules which you must follow in order to avoid a

revolt:

1. The most junior henchman (with the least seniority) gets exactly 1 LAMB. (There will always be at least 1 henchman on a team.)

2. A henchman will revolt if the person who ranks immediately above them gets more than double the number of LAMBs they do.

3. A henchman will revolt if the amount of LAMBs given to their next two subordinates combined is more than the number of LAMBs they

get. (Note that the two most junior henchmen won’t have two

subordinates, so this rule doesn’t apply to them. The 2nd most junior

henchman would require at least as many LAMBs as the most junior

henchman.)

4. You can always find more henchmen to pay – the Commander has plenty of employees. If there are enough LAMBs left over such that

another henchman could be added as the most senior while obeying the

other rules, you must always add and pay that henchman.Note that you may not be able to hand out all the LAMBs. A single LAMB

cannot be subdivided. That is, all henchmen must get a positive

integer number of LAMBs.Write a function called answer(total_lambs), where total_lambs is the

integer number of LAMBs in the handout you are trying to divide. It

should return an integer which represents the difference between the

minimum and maximum number of henchmen who can share the LAMBs (that

is, being as generous as possible to those you pay and as stingy as

possible, respectively) while still obeying all of the above rules to

avoid a revolt. For instance, if you had 10 LAMBs and were as

generous as possible, you could only pay 3 henchmen (1, 2, and 4

LAMBs, in order of ascending seniority), whereas if you were as stingy

as possible, you could pay 4 henchmen (1, 1, 2, and 3 LAMBs).

Therefore, answer(10) should return 4-3 = 1.To keep things interesting, Commander Lambda varies the sizes of the

Lucky LAMB payouts: you can expect total_lambs to always be between 10

and 1 billion (10 ^ 9).## Languages

To provide a Python solution, edit solution.py To provide a Java

solution, edit solution.java## Test cases

Inputs:

(int) total_lambs = 10 Output:

(int) 1Inputs:

(int) total_lambs = 143 Output:

(int) 3Use verify [file] to test your solution and see how it does. When you

are finished editing your code, use submit [file] to submit your

answer. If your solution passes the test cases, it will be removed

from your home folder.

```
def answer(lambs):
totals = [generous(lambs), stingy(lambs)]
difference = max(totals) - min(totals)
return difference
def generous(lambs):
num = 1
while True:
total = 2**(num) - 1
if total <= lambs:
num += 1
else:
num -= 1
break
return num
def stingy(lambs):
num = 1
while True:
total = (fibonacci(num + 2) - 1)
if total <= lambs:
num += 1
else:
num -= 1
break
return num
def fibonacci(num):
if num > -1 and num < 3:
return 1
else:
num = fibonacci(num - 1) + fibonacci(num - 2)
return num
```

Solution

Short answer: don’t use recursion.

Slightly longer answer: don’t use recursion if you have more than a single recursive call of your function in the non-trivial paths without memoization.

Your `fibonacci`

calls itself twice, except when `num`

is in `range(0,3)`

. Those two calls will call `fibonacci`

again twice (unless one of them is already in your base case), so we now end up with four calls. Let’s have a look at `fibonacci(5)`

for example:

```
fibonacci(5) = fibonacci(5 - 1) + fibonacci(5 - 2)
= (fibonacci((5 - 1) - 1) + fibonacci((5 - 1) - 2))
+ (fibonacci((5 - 2) - 1) + fibonacci((5 - 2) - 2))
= (fibonacci(((5 - 1) - 1) - 1) + fibonacci(((5 - 1) - 1) - 2)
+ fibonacci((5 - 1) - 2))
)
+ (fibonacci((5 - 2) - 1) + fibonacci((5 - 2) - 2))
= fibonacci(2) + fibonacci(1) + fibonacci(2)
+ fibonacci(2) + fibonacci(1)
```

We end up with five calls of the base case, which is what you would expect. But if we count the in-between calls, we get `9`

calls for `num = 5`

, `15`

calls for `num = 6`

, `25`

calls for `num = 7`

. The number of calls increases by a factor of 1.618$1.618$.

This is called an *exponential time algorithm*, because its runtime has the asymptotic behaviour of O(1.618n)$\mathcal{O}({1.618}^{n})$. The recursive variant looks great on paper, and it’s how Fibonacci numbers are defined after all, but the run-time is ungodly.

One can speed the recursive variant up if we would remember the values; if we calculate `fib(5)`

, we can remember the result of `fib(3)`

that we need in `fib(4) = fib(3) + fib(2)`

for our later use in `fib(5) = fib(4) + fib(3)`

. That process is called memoization, but it’s an overkill and not easy to implement.

Instead, write `fibonacci`

in an iterative fashion:

```
def fibonacci(num):
""""Returns the num-th fibonacci number, 1-indexed.
The Fibonacci sequence follows the following rules:
1) The first number is 1
2) The second number is 1
3) Every other number is the sum of its two predecessors
This yields the sequence
1,1,2,3,5,8,13,21,34,...
"""
a,b = 1,1
for _ in range(0,num):
a, b = b, a + b
return a
```

Other than that, your code seems fine. You could use better names and documentation, but given that you write code for a code challenge, it’s not that necessary. You might want to use a binary search for your limits, though, if you need more speed.

In `generous`

, you can speed things up if you use bitshifts instead of `2**num`

.

You can speed up `stingy`

if you calculate the Fibonacci numbers there. That prevents you from calculating the same sequence over and over again. Hint: you can implement `fibonacci`

as a generator, which makes this a lot easier.

Building upon @Zeta excellent answer, if you turn `fibonacci`

into a generator, you only need to sum each term in `stingy`

and stop when the sum is higher than the amount of lambs. The number of sums being the number of henchmen you can hire. To do so efficiently, you can use `itertools.accumulate`

.

You are also reinventing the wheel in `generous`

as what you are computing is a base 2 logarithm.

Lastly, it’s pretty clear in the question that the stingy strategy will always give you more henchmen than the generous one. So there is no need for `min`

and `max`

here. But even if you want to be sure, you’d rather use `abs`

instead.

Proposed improvements:

```
from math import log2
from itertools import accumulate
def answer(lambs):
return stingy(lambs) - generous(lambs)
def generous(lambs):
return int(log2(lambs + 1))
def stingy(lambs):
for henchmen, total_pay in enumerate(accumulate(fibonacci())):
if total_pay > lambs:
return henchmen
def fibonacci():
a, b = 1, 1
while True:
yield a
a, b = b, a + b
```