You have 2 identical glass bulbs with you. Bulb
manufacturer has mentioned that each bulb might withstand a
drop of 200 Feet at maximum. Your task is to find the
height at which the bulb breaks ofcourse with minimum
number of iterations.
Assume that you have 200 blocks of 1 foot each which can be
stacked one by one to create a 200 Feet structure to carry
out the test.
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Answer / chris uzdavinis
Taking the worst case scenario, the minimum number of
iterations required is 20. Here is why:
For N drops you can cover a range of the sum(1,..N), which
is n*(n+1) / 2.
If we have 200 potential heights to test, we solve n*(n+1) =
400, and using the quadratic equation gives us
19.5062490237426, rounding up to 20.
The sequence of tests would be drop at heights of:
20
39
57
74
90
105
119
132
144
155
165
174
182
189
195
200
(Notice, the distance between each level decreases by 1 to
account for the extra drop it took to get there.)
If the bulb does not break on a level, try the next higher
number in the above sequence. If it does break, then do a
linear search starting at the previous "level", plus 1.
Any answer that partitions segments into equal lengths will
be suboptimal.
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / ram kumar.y
make the stack into 14 sections such that 10 of them will
have 14 blocks each and rest of the 4 will have 15 blocks
each.i.e., 14*10+15*4=140+60=200 blocks.
coming from the bottom keep dropping the bulb from the
highest blocks in each sections. in the worst case scenario,
the bulb may break in last(14th) section.
with the second bulb, keep dropping the bulb from any one
end of that section(in this case 14th as its the worst case)
which has 15 blocks and the first bulb has been dropped from
one of the 15 blocks. so, it takes 14 more chances(at a
worst case) to find out the block from where the bulb breaks
when dropped. So, the total no. of iterations are 14+14=28(max).
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / shantanu_sanjeev
For the best solution the worst case max drops will be "24"
For this we divide it into
1st five blocks from bottom of size 20 : 5*20 = 100
Next ten blocks of top of size 10 : 10*10 = 100
This will have a worst case at 99 when there will be 24
drops and all other numbers above 99 will have less than 24
drops.
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / magnum
in the question it is asked "THE HEIGHT AT WHICH BULB BREAKS"
not the height from which u drop so the bulb breaks.so from whatever height u may drop, the bulb if it breaks,it will break only on the ground i.e. 0 feet.
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / md
200 feet is the max height.
Means bulb can break from any feet within the max limit.
In the worst case, no of iteration will be 200.
To make situation better.. lets segmentise the 200 feet in
X segments.
Then will experiment with one bulb from top of each
segments. Bulb may break at any trial. worst case bulb
breaks at last trial. iteration = X
Now need to find the exact feet among the particular
segment. This segment has 200/X feet span.
Hence Worst case : iteration I = X + (200/X)
Now the job is to find out value of X so that I is minimum.
X=10 or X=20 in worst case min no of iteration is 30
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / chris uzdavinis
Minor typo in previous answer (#7): it should say ... we
solve n*(n+1) = 200
To help get the mindset of the answer, think of the question
differently at first. Rather than solve for the fewest
number of drops to cover a range, think about a twisted
version of that question: "given a fixed number of drops,
what range can be completely covered?"
Our constraints are we have two bulbs, and we're done once
the 2nd breaks, but if it doesn't break we can use it again.
Given only 1 drop, we can only cover a range of 1. Either
it breaks or it doesn't. (If we go higher than the first
level, we don't know if it would have broken at a lower level.)
Given 2 drops, we can cover a range of 3: first drop on 2,
if it breaks, try 1, if not, try 3.
Given 3 drops, we cover a range 1..6: First drop on 3, if
it breaks, try 1, then 2. If it doesn't break, drop #2 is
on 5. If breaks, try 4. If not, try 6.
Given 4 drops, we cover a range of 1..10:
Start on 4, if breaks, try 1,2,3. If not, try 7. If
breaks, try 5,6. If not, try 9. If breaks, try 8. If not
try 10.
And so on. Thus, for N drops, we start on level N. If it
breaks, start a linear search using the remaining bulb, from
1..N-1. Thus, it took a max of 20 drops. But if it doesn't
break, we try level N+(N-1). If it breaks, start a linear
search from N+1. Otherwise try N+(N-1)+(N-2). etc.
This series is clearly the sum from 1 to N drops.
So now that we know how big of a range we can cover given a
fixed number of drops. But the original question asks us to
cover a range of 200, and find the number of drops required.
We just find the smallest integral value for N such that
sum of 1..N >= 200, which is 20.
I gave the sequence of tests in the previous post.
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / vineeta agarwal
Max height = 200 feet
no of bulbs we have=2
in minimum iteration we need to find the height at which bulb will break...
i think we can find only range... not exact height...
first will divide 200/2=100 feet
we drop bulb from 100 feet if 1st bulb break than we have to find height from 0 to 100 feet.... than we drop 2nd bulb from 50 feet if break than our range is 0-50 else we do same thing from 75 feet..... just like that we do and will find the range....
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / viral avalani
No. of iterations will be minimum at height of 200 ft.
| Is This Answer Correct ? | 0 Yes | 1 No |
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