## Oracle Interview Question

Senior Software Development Engineers**Country:**India

**Interview Type:**In-Person

there are a few ways to go about it..

1 split it in half (equal parts) and weigh both against each other,

which half has lower weight has the defective coin,

repeat until you have only one left

2 split it in 4 or 3 (equal) parts compare 2,

if one has lower the defective coin is in that part,

if both weigh the same the defective part is in

the 2 or 1 part that was not compared

usually explain the first part, and if they say this can be optimised then explain the second method(pause enough so that your answer seems legit ðŸ˜Š)

Place three coins on each arm of the scale, leaving the two coins out.

If the scales tip, take the three lighter coins and discard the rest. If the scales don't tip, the defective coin was not on the scale so take the two remaining coins and discard the rest. Place one coin on each arm of the scale, leaving a third coin out if the scales previously tipped.

If the scales tip on the second weigh-in, the defective coin is the lighter of the two. If the scales don't tip, the coin is the one remaining coin.

This method guarantees you use the scales a minimum and maximum of two times.

Divide the balls into two parts as 6 balls and 2 balls

Take 3 and 3 from first part and wighs.

If

the weighs are equal then take other part which is having 2 balls and weigh again which ever is lesser wieghs that will be the faulty one.

else

take the three balls set weighs lesser in the first turn.

Take two balls from the set 3 and wiegh ,

if both wieghs same, left over ball is the faulty one.

else

which ever ball is lesser wighs that will be faulty one.

End

In this way we can identify the faulty ball in only 2 Iterations.

there are a few ways to go about it..

- Anonymous July 04, 20181 split it in half (equal parts) and weigh both against each other,

which half has lower weight has the defective coin,

repeat until you have only one left

2 split it in 4 or 3 (equal) parts compare 2,

if one has lower the defective coin is in that part,

if both weigh the same the defective part is in

the 2 or 1 part that was not compared

usually explain the first part, and if they say this can be optimised then explain the second method(pause enough so that your answer seems legit ðŸ˜Š)