Oracle Interview Question Software Engineer / Developers


Team: Java Developers
Country: United States
Interview Type: In-Person


Comment hidden because of low score. Click to expand.
10
of 12 vote

It can be done in 2 measures:
Divide the 8 balls in 3 sets of 3,3,2.
Measure1: Weigh Set1 against Set2 => From here you can determine whether the heavier one belongs to Set1 or Set2 or Set3(if weights are equal).
Measure2: If you are left with Set3, measure the two balls against each other. Else if you are left with Set1/Set2, weigh two of the balls from that set against each other and you can determine whether one of them is heavier or the 3rd is actually heavier.

- -- on March 14, 2012 | Flag Reply
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0
of 0 votes

suppose the wieght is 1,2,3,4,5,6,7,8,
and set is s1{1,8} s2{2,3,4} s3{6,7,8}

then s1 is eleminated which have heavier weight 8

- Anonymous on June 20, 2012 | Flag
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0
of 0 votes

The solution stated at the top is correct. Out of the 8 balls, ONLY ONE of them is heavier.

- Sandeep Mathias on July 31, 2012 | Flag
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0
of 2 vote

3 comparisons.

4,4 ==> 2,2 ==> 1,1

- Anonymous on March 14, 2012 | Flag Reply
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0
of 0 vote

Nayan is correct...it can be done in 2 steps..

- Anshul on March 14, 2012 | Flag Reply
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0
of 0 vote

with 2 steps it can be done... As nayan answer is correct

- kbhaskarkotha on March 14, 2012 | Flag Reply
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0
of 0 vote

Nayan is correct

- anonymous on March 16, 2012 | Flag Reply
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0
of 0 vote

Nayan's answer is fine, but pls clarify my doubt. Lets say we are left with set1/set2. So, in measure2, what if the two balls are of same weight in either of the sets. Then we need take that ball in that set and do one more comparison with the 3rd ball to find the heavier one right?
Hope it make sense. So, is it not totally 3 steps?

- San on March 18, 2012 | Flag Reply
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0
of 0 votes

In case u happen to select 2 balls from set 1/set2 which weigh the same,the third ball left in that set is the heavier one,so nayan is correct

- Anonymous on March 18, 2012 | Flag
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0
of 0 votes

it can be done in two measures
lets divide balls as (3 3 2)
take set (3,3)
case a) either balls are equal ( implies remaining 2 contains the heavier one)
case b) not equal,then take the heavier lot of 3 balls and divide them as (1,1,1). You can the heavier ball in just one step

so it is just 2 steps

case i)repeat the process as steps a) and b)

- Raman on April 06, 2012 | Flag
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0
of 0 votes

in (1,1,1) if first and second are compared and if one is heavier then there should be another comparision vt third ball right??

- Anonymous on May 06, 2012 | Flag
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0
of 0 votes

No need bro. If one of any two balls in a set is heavier than the other it would implicitly mean that, the ball in picture is the bulkiest one.

- Pavan Dittakavi on May 08, 2012 | Flag
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0
of 0 vote

Its absolutely in 2 comp as stated in first solution...
s1(1,1,1),s2(1,1,2),s3(1,1)
compare s1 and s2(COMP-1)
s2 is big so compare s2[0]and s2[1](COMP-) both equal so s2[2] is hevier

- NAX on June 04, 2013 | Flag Reply
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-1
of 1 vote

Divide and conquer
2^3 - so within 3 measures

- whizz.comp on March 14, 2012 | Flag Reply
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0
of 0 votes

no brother it can be done in 2 time measurement...

- abh007 on July 26, 2012 | Flag


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