This community-built FAQ covers the “Comparative Weights” code challenge in Python. You can find that challenge here, or pick any challenge you like from our list.
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I keep getting only 4/5 tests passed, i just can’t figure out what is missing on my code, for me it works 100%.
Can anyone please point out what i am missing?
c = 0
while n != 1:
n = n//2
c += 1
if n < 1:
count = 0
while n != 1:
while n % 3 != 0:
n += 1
n /= 3
count += 1
This code passed all the tests
if you put in 0 you would get an endless loop thats probably why. But i think your method is wrong, for 81 as an example you split it into 3 groups, 27 now, then again, 9 now, then again,3 now and finally once more to get the one. thats 4 times whereas your code would say 8.
read the prompt again to see why its 3 groups not 2
if n < 2: return 0
return math.ceil(math.log(n, 3))
Can’t be too creative here, just handing in my own solution which is basically identical to what @text9831554157 's most recently posted for the reasons @text9831554157 posted before.
yeah this one the math is really the solution
Could someone please explain why I have to use log with a base of 3 instead of base 2? It seemed more logical to me to divide the number of N by 2 until I reached a setting in which I would have only 1 ball on either side of the scale. Consequently, the number of weighs would be math.ceil(math.log(n,2)).
log base 2 works perfectly fine too, it’s just not the fastest solution. The trick with this challenge is:
- From your n balls, form three piles of equal size.
- Put pile 1 on the one side of your balance scale, pile 2 on the other side, and pile 3 aside.
- If the scales tilt to either side, you know that the heavier ball is on that side.
- If the scales remain balanced, you know that the heavier ball has to be in your third pile
- With the identified pile: repeat until there’s only one ball left in your identified pile
This way you can cut the remaining number of balls in thirds with every weighing, hence the log base 3.
from math import floor, log
if n==1: return 0
x = floor(log(n-1,3))+1