Oracle Interview Question for Senior Software Development Engineers


Country: India
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
0
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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 😊)

- Anonymous July 04, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

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 😊)

- PeyarTheriyaa July 04, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

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.

- Anonymous August 05, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

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.

- sbhargavs August 11, 2018 | Flag Reply


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