CareerCup

I have no idea what this question is, can you provide more details on the question?

Actually I found the problem statement which I believe is the same problem asked here:

8 ball problem. if given 8 balls with only one ball being a 
lighter weight and all other 7 being the same weight and 
given a scale, how will you find the light weight ball. do 
this in minimum number of steps

My initial thought is to divide an conquer which gives

log(N)/log(2)

so in case of 8-ball, it would be

log(8)/log(2) = 3 steps

, but you mentioned it can be done in two steps.

The above would be true depending on the meaning of 'scale'.

- If scale allows you to weigh one group at a time( ex Think of weighing yourself on scale, only one person at a time makes sense ), then I believe

log(8)/log(2)

would be true.

- If scale allows you to weight two group at a time( another type of scale ) then (after cheating and reading answer (see /question?id=8119690) ) they mention the answer is

ceil(log(N)/log(3))

But its only that because they skip the last ball weigh-in using slick process of elimination. So technically they don't weight all the balls.

While this problem is interesting, I'm not a fan of it, especially for a programming interview position( assuming ).

Best.

*cough *cough * I hate this site broken functionalities *cough *cough

public int getHeavy(List<Integer> balls)
{
	return (int)Math.ceil(Math.log(balls.size())/Math.log(3));
{