I still like my answer. If you luck out and pick the heavy ball and a regular ball on your first weigh-in, then that lucky pick would represent the absolute minimum number of attempts to find the heavy ball. I'm sure I'm probably missing something in this scenario, but that's my logic... for now.
No No No;seriously? IE., the intent is a reliable solution, one that can be counted on to give a correct result every time. You don't get that?
0 If they're identical but one is heavier, then it must be denser. Find a fluid thick enough where they kind of float, and the one that floats to the lowest point is the heaviest one. No need for scale. If they are of equal density, then the odd one must be a different size. Look for the biggest one. Again 0 because you don't need a scale. I never like these "out of the box" interview questions, because the person judging you will compare your answer against his googled answer (definitely not out of the box). If you give a truly original answer like some posted here, then you are faulted for a 'wrong' answer.
"Suppose you had eight identical balls. One of them is slightly heavier and you are given a balance scale. What's the fewest number of times you have to use the scale to find the heavier ball EVERYTIME?" There i fixed it
try this one Baron (its old, maybe you've seen it before)but a good one: The puzzle: You are presented with 3 bags of gold. One of the bags has "fake" gold in it. The fake gold is heavier than real gold. You have a penny scale and only one penny with which to purchase a weighing. You cannot add or remove pieces during the weighing -- one penny = one reading. How do you identify the bag with the fake gold?
Finding the balls answer is 3. Divide balls in half, weigh 4 on each scale. Divide the heavy 4 in half, then weight. Take the heaviest 2, divide and weigh. Heavy ball found, every time. I got big balls on that answer, no google, just big brain