Math/CompSci Interview Question - Thoughts?

[Andrew Tomazos]

I was posed the following question in a technical interview for a
Software Engineering position by a major multinational NASDAQ company:

[Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
One int value x occurs 500,001 times or more in the array. Specify an
algorithm to determine x."

The assumption being extra points for efficiency.

I initially came up with a linear space, linear time solution. With
some prompting and hints from the interviewer we worked our way to a
smaller constant space and linear time algorithm. That being
basically:

int findmode(int* p, int n)
{
int count[32];
for(int i = 0; i < 32; i++)
count = 0;

for (int i = 0; i < n; i++)
for (int j = 0; j < 32; j++)
if (p & (1 << j)) // if bit j is on
count[j]++;
else
count[j]--;

int x = 0;
for (int i = 0; i < 32; i++)
if (count > 0)
x = x | (1 << i);

return x;
}

The idea here is to count the frequency of each of the 32 bits in the
array separately, knowing that these bit counts will be dominated by
the mode.

The interviewer already knew the answer, so I can only assume the test
was based on how much help he had to give me to arrive at it.

My question about this is as follows. I, perhaps boldly, claim to
employers to have a "masters-equivilant" experience/knowledge of
computer science and math. Should I have been able to come up with
this solution without prompting or help?

Would you expect someone with a CompSci Masters or PhD from some major
ivy league university to be able to come up with this solution without
help?

If so in what course or from which book would you expect to learn the
required knowledge to come up with the above solution?

Is the skill to be able to come up with such an algorithm something
that is learned by studying lots of algorithms and being able to mix
and match the "tricks"? If so, what is a similar commonly known
algorithm(s) on which the above solution could have been based?

Or, is the ability to invent such a solution simply a matter of IQ?
Some people have the talent to solve the puzzle, see the pattern and
come up with the solution - and some just don't?

Thanks,
Andrew.

 

[Pascal J. Bourguignon]

I was posed the following question in a technical interview for a
Software Engineering position by a major multinational NASDAQ company:

[Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
One int value x occurs 500,001 times or more in the array. Specify an
algorithm to determine x."

The assumption being extra points for efficiency.

I initially came up with a linear space, linear time solution. With
some prompting and hints from the interviewer we worked our way to a
smaller constant space and linear time algorithm. That being
basically:

int findmode(int* p, int n)
{
int count[32];
for(int i = 0; i < 32; i++)
count = 0;

for (int i = 0; i < n; i++)
for (int j = 0; j < 32; j++)
if (p & (1 << j)) // if bit j is on
count[j]++;
else
count[j]--;

int x = 0;
for (int i = 0; i < 32; i++)
if (count > 0)
x = x | (1 << i);

return x;
}

The idea here is to count the frequency of each of the 32 bits in the
array separately, knowing that these bit counts will be dominated by
the mode.

The interviewer already knew the answer, so I can only assume the test
was based on how much help he had to give me to arrive at it.

My question about this is as follows. I, perhaps boldly, claim to
employers to have a "masters-equivalant" experience/knowledge of
computer science and math. Should I have been able to come up with
this solution without prompting or help?

If what you're asking is whether anybody having a master in CS and
maths would have been able to come up with this solution in the
interview time, I guess we can answer that definitely no, otherwise
there would be no point in trying to select candidates with this test.

Obviously, it's because some people (with or without a master diploma,
this really isn't relevant) get or don't get it, that this test is
useful for the recruiter.



Now if you want this kind of jobs, yes you should better be able to
come up with smart solutions to little puzzles like this in
interviews.

Would you expect someone with a CompSci Masters or PhD from some major
ivy league university to be able to come up with this solution without
help?

If so in what course or from which book would you expect to learn the
required knowledge to come up with the above solution?

Not a single one. You have to develop your knowledge of algorithms,
mathematics, your symbolic thinking and your imagination in these
matters.

Is the skill to be able to come up with such an algorithm something
that is learned by studying lots of algorithms and being able to mix
and match the "tricks"?

That could help yes. I'd tend to think that imagination is the most
important ingredient here, to be able to come with a possible solution
fast enough.

Also, don't limit yourself to CS and maths, there are ideas to be
taken from other domains too.

If so, what is a similar commonly known
algorithm(s) on which the above solution could have been based?

Or, is the ability to invent such a solution simply a matter of IQ?

CS IQ, yes, if such a IQ test exists.

Some people have the talent to solve the puzzle, see the pattern and
come up with the solution - and some just don't?

It seems so. At least, in a given time.

 

[GJ Woeginger]

# I was posed the following question in a technical interview for a
# Software Engineering position by a major multinational NASDAQ company:
#
# [Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
# One int value x occurs 500,001 times or more in the array. Specify an
# algorithm to determine x."
#
# The assumption being extra points for efficiency.

There is an old analysis of this problem by Fischer and Salzberg.
M.J. Fisher and S.L. Salzberg (1982)
Finding a majority among n votes.
Journal of Algorithms 3, pp 143-152.

If 2k elements contain a majority element (= an element that occurs at
least k+1 times), then you can find it with 3k-2 element comparisons
(and some small overhead). The basic idea in their algorithm is that
whenever you find two distinct elements, then you can destroy both without
changing the majority element among the remaining elements. This yields:

Run once through the array, and maintain a majority-candidate and a counter.
The majority-candidate is initialized as the first element, and
the counter (counting how often you have seen the candidate) is
initialized at 1.
If you hit the current candidate again, increment the counter.
If you hit another element, decrement the counter.
If the counter becomes 0, drop the current candidate and start from
scratch with the remaining part of the array.

Fischer and Salzberg also show that this bound 3k-2 is best possible in
the worst case (and that's the main part of their paper).

--Gerhard

___________________________________________________________
Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/

 

[A.G.McDowell]

I was posed the following question in a technical interview for a
Software Engineering position by a major multinational NASDAQ company:

[Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
One int value x occurs 500,001 times or more in the array. Specify an
algorithm to determine x."

The assumption being extra points for efficiency.

I initially came up with a linear space, linear time solution. With
some prompting and hints from the interviewer we worked our way to a
smaller constant space and linear time algorithm. That being
basically:

This sort of problem is covered by articles on Data Stream Processing in
CACM Oct 2009. (CACM is a lot more interesting these days than it was
some years ago). There are some very neat ideas in there, of which the
algorithm "MAJORITY" matches the question reasonably well. Proving that
it works under interview conditions would be extremely impressive,
though.

This is not the first time that I have heard of interview questions that
discuss issues recently covered in the computing literature. I am unable
to tell whether these come from a desire to know if the candidate keeps
themselves abreast of the subject, or from the interviewer grasping the
first thing that comes to hand when they are trying to think up a
question. The few times that I have posed interview questions, I have
tried to find evidence in the candidate of a knowledge of basic theory
or mathematics that I could show was relevant to the job for which we
were trying to recruit.

 

[Bill Dubuque]

I was posed the following question in a technical interview for a
Software Engineering position by a major multinational NASDAQ company:

[Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
One int value x occurs 500,001 times or more in the array. Specify an
algorithm to determine x."

The assumption being extra points for efficiency. [snip code] The idea
is to count the frequency of each of the 32 bits in the array separately,
knowing that these bit counts will be dominated by the mode. [...]

My question about this is as follows. I, perhaps boldly, claim to
employers to have a "masters-equivilant" experience/knowledge of
computer science and math. Should I have been able to come up with
this solution without prompting or help?
Would you expect someone with a CompSci Masters or PhD from some major
ivy league university to be able to come up with this solution without
help? If so in what course or from which book would you expect to learn
the required knowledge to come up with the above solution?

That's an interesting question with an interesting presupposition.
The first thing to understand is that this is a puzzle rather than
a programming problem.

I disagree. Saying that it's a puzzle rather than a programming problem
seems to imply that the solution is ad-hoc, rather than a special case
of some more universal technique. But that is not true here. Namely,
the proposed solution follows immediately from the obvious fact that
majority elts on product sets are preserved by component projections.
So the solution is simply an instance of well-known divide-and-conquer
techniques for product objects. Such reductions are ubiquitous in both
mathematics and computer science. So I would expect a good student
to find this solution given enough time. I'd also expect a student
from a top-tier school to discover the more optimal solution, viz.

bi != a => Maj({a^k b1 b2... bk} \/ S) = Maj(S)

via m/n > 1/2 => (m-k)/(n-2k) > 1/2 via mediants/arithmetic

again, assuming enough time. But that assumption is highly problematic
when it comes to designing tests that quickly measure various skills.
It's impossible to test such in the short time span of an interview.
It remains difficult even in much longer timed tests. For example
many of the most successful mathematicians did not perform well on
the Putnam competition, and some of those who did well didn't turn
out to be exceptional mathematicians. Thus there isn't necessarily
much correlation between intelligence and random test scores.

Marilyn vos Savant is a prime example. The woman who supposedly
has the "world's highest IQ" published a Parade magazine column [1]
and book [2] on Wiles proof of FLT. Her nonsensical arguments proved
beyond a doubt that she has absolutely no clue about higher math
and, worse, doesn't have the intelligence to know that (even after
many experts pointed out gaping flaws in her very naive arguments).
So much for IQ tests.

--Bill Dubuque

[1] sci.math, 11/29/1993, Full Text of Savant FLT Parade Article
http://google.com/group/sci.math/msg/8cb44534c63372ad
http://google.com/[email protected]

[2] Boston, Nigel; Granville, Andrew.
Review of: The World's Most Famous Math Problem (The Proof
of Fermat's Last Theorem and Other Mathematical Mysteries)
by Marilyn vos Savant
The American Mathematical Monthly, 102, 5, 1995, 470-473
http://www.dms.umontreal.ca/~andrew/PDF/VS.pdf
http://www.jstor.org/stable/2975048

 

[Alf P. Steinbach]

* Bill Dubuque:

I disagree. Saying that it's a puzzle rather than a programming problem
seems to imply that the solution is ad-hoc, rather than a special case
of some more universal technique. But that is not true here. Namely,
the proposed solution follows immediately from the obvious fact that
majority elts on product sets are preserved by component projections.
So the solution is simply an instance of well-known divide-and-conquer
techniques for product objects. Such reductions are ubiquitous in both
mathematics and computer science. So I would expect a good student
to find this solution given enough time. I'd also expect a student
from a top-tier school to discover the more optimal solution, viz.

bi != a => Maj({a^k b1 b2... bk} \/ S) = Maj(S)

via m/n > 1/2 => (m-k)/(n-2k) > 1/2 via mediants/arithmetic

Hiya. Could you please explain the above more optimal solution. I'm unfamiliar
with the notation and not from a top-tier school, but in my experience anything
that's not nonsense can be visualized or explained in simple terms (e.g., Albert
Einstein did that beautifully with his book on special relativity, which except
for one little proof in an appendix -- I didn't yet know anything about
solving sets of equations with multiple variables -- I found eminently
grokkable as a teenager, so should also be possible for the above, yes?).

Cheers & TIA.,

- Alf

 

[Axel Vogt]

I was posed the following question in a technical interview for a
Software Engineering position by a major multinational NASDAQ company:

[Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
One int value x occurs 500,001 times or more in the array. Specify an
algorithm to determine x."

The assumption being extra points for efficiency.

<snipped>

Being a bit stupid I first would ask "Why? What do you do with it?"
and then I would pick on random. I am almost certain, that even at
a low number of draws the chance to get the very integer is higher
than implementing an algo without coding errors.

 

[Andrew Tomazos]

* Bill Dubuque:






Hiya. Could you please explain the above more optimal solution.

Dubuque is referring to the solution that Woeginger more lucidly
described above. Both it and the bit counting method are
asymptotically equivalent solutions to the original problem. I'm sure
either of these solutions provided on the spot would have received
"full marks".

I guess what I am curious about is exactly what percentage of, say...
CS PhD students at tier one universities, would be able to come up
with either of these solutions on the spot. 80%, 50% or 20% ? I
guess only someone that has interviewed many people with these sort of
problems has the necessary data to answer my question.
-Andrew.

 

[Andrew Tomazos]

# I was posed the following question in a technical interview for a
# Software Engineering position by a major multinational NASDAQ company:
#
# [Paraphrasing]  "You are given an array of 1,000,000 32-bit integers.
# One int value x occurs 500,001 times or more in the array.  Specify an
# algorithm to determine x."
#
# The assumption being extra points for efficiency.

There is an old analysis of this problem by Fischer and Salzberg.
  M.J. Fisher and S.L. Salzberg  (1982)
  Finding a majority among n votes.
  Journal of Algorithms 3, pp 143-152.

If 2k elements contain a majority element (= an element that occurs at
least k+1 times), then you can find it with 3k-2 element comparisons
(and some small overhead).  The basic idea in their algorithm is that
whenever you find two distinct elements, then you can destroy both without
changing the majority element among the remaining elements.  This yields:

 Run once through the array, and maintain a majority-candidate and a counter.
 The majority-candidate is initialized as the first element, and
   the counter (counting how often you have seen the candidate) is
   initialized at 1.
 If you hit the current candidate again, increment the counter.
 If you hit another element, decrement the counter.
 If the counter becomes 0, drop the current candidate and start from
   scratch with the remaining part of the array.

Fischer and Salzberg also show that this bound 3k-2 is best possible in
the worst case (and that's the main part of their paper).

If I understand your description than it would look like:

int findmode(int* p, int n)
{
int x = p[0];
int c = 1;

for (int i = 1; i < n; i++)
{
if (c == 0)
{
x = p;
c = 1;
}
else if (p == x)
c++;
else
c--;
}

return x;
}

It seems that this could only produce at maximum (2k-1) comparisions
in the worst case, not (3k-2) as you claim?
-Andrew.

 

[Andrew Tomazos]

# I was posed the following question in a technical interview for a
# Software Engineering position by a major multinational NASDAQ company:
#
# [Paraphrasing]  "You are given an array of 1,000,000 32-bit integers.
# One int value x occurs 500,001 times or more in the array.  Specify an
# algorithm to determine x."
#
# The assumption being extra points for efficiency.

There is an old analysis of this problem by Fischer and Salzberg.
  M.J. Fisher and S.L. Salzberg  (1982)
  Finding a majority among n votes.
  Journal of Algorithms 3, pp 143-152.

If 2k elements contain a majority element (= an element that occurs at
least k+1 times), then you can find it with 3k-2 element comparisons
(and some small overhead).  The basic idea in their algorithm is that
whenever you find two distinct elements, then you can destroy both without
changing the majority element among the remaining elements.  This yields:

 Run once through the array, and maintain a majority-candidate and a counter.
 The majority-candidate is initialized as the first element, and
   the counter (counting how often you have seen the candidate) is
   initialized at 1.
 If you hit the current candidate again, increment the counter.
 If you hit another element, decrement the counter.
 If the counter becomes 0, drop the current candidate and start from
   scratch with the remaining part of the array.

Fischer and Salzberg also show that this bound 3k-2 is best possible in
the worst case (and that's the main part of their paper).

If I understand your description than it would look like:

int findmode(int* p, int n)
{
    int x = p[0];
    int c = 1;

    for (int i = 1; i < n; i++)
    {
       if (c == 0)
       {
          x = p;
          c = 1;
       }
       else if (p == x)
           c++;
       else
           c--;
    }

    return x;

}

It seems that this could only produce at maximum (2k-1) comparisions
in the worst case, not (3k-2) as you claim?
  -Andrew.

Oh wait... the (c==0) counts as a comparison. I see it now. Ignore
me.
-Andrew.

 

[Bill Dubuque]

(e-mail address removed) (Richard Harter) wrote:

I was posed the following question in a technical interview for a
Software Engineering position by a major multinational NASDAQ company:

[Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
One int value x occurs 500,001 times or more in the array. Specify an
algorithm to determine x."

The assumption being extra points for efficiency. [snip code] The idea
is to count the frequency of each of the 32 bits in the array separately,
knowing that these bit counts will be dominated by the mode. [...]

My question about this is as follows. I, perhaps boldly, claim to
employers to have a "masters-equivilant" experience/knowledge of
computer science and math. Should I have been able to come up with
this solution without prompting or help?
Would you expect someone with a CompSci Masters or PhD from some major
ivy league university to be able to come up with this solution without
help? If so in what course or from which book would you expect to learn
the required knowledge to come up with the above solution?

That's an interesting question with an interesting presupposition.
The first thing to understand is that this is a puzzle rather than
a programming problem.

I disagree. Saying that it's a puzzle rather than a programming problem
seems to imply that the solution is ad-hoc, rather than a special case
of some more universal technique. But that is not true here. Namely,
the proposed solution follows immediately from the obvious fact that
majority elts on product sets are preserved by component projections.
So the solution is simply an instance of well-known divide-and-conquer
techniques for product objects. Such reductions are ubiquitous in both
mathematics and computer science. So I would expect a good student
to find this solution given enough time. I'd also expect a student
from a top-tier school to discover the more optimal solution, viz.

bi != a => Maj({a^k b1 b2... bk} \/ S) = Maj(S)

via m/n > 1/2 => (m-k)/(n-2k) > 1/2 via mediants/arithmetic

again, assuming enough time. But that assumption is highly problematic
when it comes to designing tests that quickly measure various skills.
It's impossible to test such in the short time span of an interview.
It remains difficult even in much longer timed tests. For example
many of the most successful mathematicians did not perform well on
the Putnam competition, and some of those who did well didn't turn
out to be exceptional mathematicians. Thus there isn't necessarily
much correlation between intelligence and random test scores.

Marilyn vos Savant is a prime example. The woman who supposedly
has the "world's highest IQ" published a Parade magazine column [1]
and book [2] on Wiles proof of FLT. Her nonsensical arguments proved
beyond a doubt that she has absolutely no clue about higher math
and, worse, doesn't have the intelligence to know that (even after
many experts pointed out gaping flaws in her very naive arguments).
So much for IQ tests.

[1] sci.math, 11/29/1993, Full Text of Savant FLT Parade Article
http://google.com/group/sci.math/msg/8cb44534c63372ad
http://google.com/[email protected]

[2] Boston, Nigel; Granville, Andrew.
Review of: The World's Most Famous Math Problem (The Proof
of Fermat's Last Theorem and Other Mathematical Mysteries)
by Marilyn vos Savant
The American Mathematical Monthly, 102, 5, 1995, 470-473
http://www.dms.umontreal.ca/~andrew/PDF/VS.pdf

Hiya. Could you please explain the above more optimal solution.

I'll be happy to elaborate if you say what isn't clear to you.

Dubuque is referring to the solution that Woeginger more lucidly
described above.

No, I hadn't yet seen Woeginger's post when I posted that. Note
that what one finds "more lucid" may depend on one's background.
A computer scientist might find W's post more lucid, whereas a
mathematician may find the above more to the essence of the matter,
esp. since the mediant viewpoint make the underlying math trivial
(W's post completely omits discussion of any math or proof).

--Bill Dubuque

 

[Casey Hawthorne]

These questions/puzzles are to see how you approach the problem, but
if you've seen the puzzle before it is a useless practice.

 

[Willem]

Virgil wrote:
) I
)> >>>I was posed the following question in a technical interview for a
)> >>>Software Engineering position by a major multinational NASDAQ company:
)> >>>
)> >>>[Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
)> >>>One int value x occurs 500,001 times or more in the array. Specify an
)> >>>algorithm to determine x."
)
) Find, to the nearest integer, the average value.

Doesn't work.
Consider the case of 500,001 times 0 and 499,999 times 2^31-1

Perhaps you are confused with 'find the median value' ?


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

 

[Willem]

Andrew Tomazos wrote:
) Dubuque is referring to the solution that Woeginger more lucidly
) described above. Both it and the bit counting method are
) asymptotically equivalent solutions to the original problem. I'm sure
) either of these solutions provided on the spot would have received
) "full marks".

I'm not so sure. I wouldn't be surprised if you only get "full marks"
for the solution that the interviewer has in mind, even if there are
other solutions that are equivalent or better.

NB: The counting method is actually better, because you don't need
O(k * log n) space (where k is the size of an individual key).
Or, practically speaking, it also works if the items are strings.


A very simple proof of the 'other' method would be that whenever
c reaches 0, the part of the array you processed can't have more
than half of the 'majority' item. (Exactly half of it is one single
item, and the other half is not that single item) So the remaining
part *must* have more than half, i.e. the problem is reduced to
the same problem for the remaining part.


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

 

[Moi]

I was posed the following question in a technical interview for a
Software Engineering position by a major multinational NASDAQ company:

[Paraphrasing] "You are given an array of 1,000,000 32-bit [integers.
One int value x occurs 500,001 times or more in the array. Specify an
algorithm to determine x."

The assumption being extra points for efficiency.

<snipped>

Being a bit stupid I first would ask "Why? What do you do with it?" and
then I would pick on random. I am almost certain, that even at a low
number of draws the chance to get the very integer is higher than
implementing an algo without coding errors.

I thought of that too, but quickly saw the flaw - if for large N you
have (N/2)+1 occurrences of X, and (N/2)-1 occurrences of Y, you can't
settle the matter (is it X, or is it Y?) satisfactorily without reading
every element.

With a kind of weight-balanced tree,
( := rotate in such a way that the node with the higher reference
count is closer to the root) you can terminate reading and inserting
the values once there is a clear "winner".

I think, something similar can be done for the bitcounting method:
if all bins have a value with (abs(value) > number_of_remaining_items)
you can terminate the inspection / insertion.



AvK

 

[Andrew Tomazos]

) Dubuque is referring to the solution that Woeginger more lucidly
) described above.  Both it and the bit counting method are
) asymptotically equivalent solutions to the original problem.  I'm sure
) either of these solutions provided on the spot would have received
) "full marks".

I'm not so sure.  I wouldn't be surprised if you only get "full marks"
for the solution that the interviewer has in mind, even if there are
other solutions that are equivalent or better.

Ummm. No, I don't think so. It was my impression that performance is
assessed according to the asymptotic equivalence class of time and
space requirements. At least in the interview setting I am talking
about.

NB: The counting method is actually better, because you don't need
O(k * log n) space (where k is the size of an individual key).
Or, practically speaking, it also works if the items are strings.

Normally, k (the number of bits in an int) would be considered
constant (I haven't encountered many 1,000,000 bit integers as yet),
therefore both solutions are asymptotically equivalent in space and
time. There is usually a very large number of ways to make
nonasymptotic improvements to any algorithm. None of this precludes
one solution being "better" than the other by some other measure
though, but it gets harder to measure performance at a finer grain.
You need to start considering the impact of the memory hierarchy,
locality of reference, what the optimizer is going to do, and so on.
To be honest I would expect both algorithms to be memory bound by the
single pass of the large array. But it would be interesting to
compare anyway. Maybe I'll do a performance comparison.
-Andrew

 

[Willem]

Andrew Tomazos wrote:
)> I'm not so sure.  I wouldn't be surprised if you only get "full marks"
)> for the solution that the interviewer has in mind, even if there are
)> other solutions that are equivalent or better.
)
) Ummm. No, I don't think so. It was my impression that performance is
) assessed according to the asymptotic equivalence class of time and
) space requirements. At least in the interview setting I am talking
) about.

My comment is based on my general impression of interviewers, not on the
way the OP worded his question. I find it more likely that an interviewer
posing such a question 'knows the trick' and wants the interviewee to find
the same trick. Also note that in this case, the interviewer goaded the
interviewee to the less general solution (see below).

)> NB: The counting method is actually better, because you don't need
)> O(k * log n) space (where k is the size of an individual key).
)> Or, practically speaking, it also works if the items are strings.

Did you read the last sentence ? You know, the one with 'strings' ?

) Normally, k (the number of bits in an int) would be considered
) constant (I haven't encountered many 1,000,000 bit integers as yet),
) therefore both solutions are asymptotically equivalent in space and
) time.

We're looking at the difference between two algorithms. With one view,
they're equivalent, with another view one of them is better. To me, that
means that that one is better overall.

Furthermore, the counting solution is more general, because it
can work on any type of item with an equivalence operator.

All of this has nothing to do with nonasymptotic behaviour.

) <snip>


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

 

[Chip Eastham]

On Nov 21, 6:29 pm, (e-mail address removed)-graz.ac.at (GJ Woeginger)
wrote:

# I was posed the following question in a technical interview for a
# Software Engineering position by a major multinational NASDAQ company:
#
# [Paraphrasing]  "You are given an array of 1,000,000 32-bit integers.
# One int value x occurs 500,001 times or more in the array.  Specify an
# algorithm to determine x."
#
# The assumption being extra points for efficiency.
There is an old analysis of this problem by Fischer and Salzberg.
  M.J. Fisher and S.L. Salzberg  (1982)
  Finding a majority among n votes.
  Journal of Algorithms 3, pp 143-152.
If 2k elements contain a majority element (= an element that occurs at
least k+1 times), then you can find it with 3k-2 element comparisons
(and some small overhead).  The basic idea in their algorithm is that
whenever you find two distinct elements, then you can destroy both without
changing the majority element among the remaining elements.  This yields:
 Run once through the array, and maintain a majority-candidate and a counter.
 The majority-candidate is initialized as the first element, and
   the counter (counting how often you have seen the candidate) is
   initialized at 1.
 If you hit the current candidate again, increment the counter.
 If you hit another element, decrement the counter.
 If the counter becomes 0, drop the current candidate and start from
   scratch with the remaining part of the array.
Fischer and Salzberg also show that this bound 3k-2 is best possible in
the worst case (and that's the main part of their paper).

If I understand your description than it would look like:

int findmode(int* p, int n)
{
    int x = p[0];
    int c = 1;

    for (int i = 1; i < n; i++)
    {
       if (c == 0)
       {
          x = p;
          c = 1;
       }
       else if (p == x)
           c++;
       else
           c--;
    }

    return x;

It seems that this could only produce at maximum (2k-1) comparisions
in the worst case, not (3k-2) as you claim?
  -Andrew.

Oh wait... the (c==0) counts as a comparison.  I see it now.  Ignore
me.
  -Andrew.

I don't think the (c==0) counts as a comparison.
I think the other k-1 comparisons are for the 2nd
part of the algorithm, when you make one more pass
through the array verifying the candidate found in
the 1st part. [You already know the index i for
one occurrence of the candidate, so only the other
k-1 indices have to be checked.]

You can see how badly I (mathtalk-ga) stumbled
through this same puzzle here:

[Google Answers: algorithms]
http://answers.google.com/answers/threadview/id/582711.html

regards, chip

 

[Ben Bacarisse]

# I was posed the following question in a technical interview for a
# Software Engineering position by a major multinational NASDAQ company:
#
# [Paraphrasing] "You are given an array of 1,000,000 32-bit integers.
# One int value x occurs 500,001 times or more in the array. Specify an
# algorithm to determine x."
#
# The assumption being extra points for efficiency.

There is an old analysis of this problem by Fischer and Salzberg.
M.J. Fisher and S.L. Salzberg (1982)
Finding a majority among n votes.
Journal of Algorithms 3, pp 143-152.

I can't track this down. Have I got the right Journal of Algorithms?

http://www.informatik.uni-trier.de/~ley/db/journals/jal/jal3.html

does not list that paper. Of course, this table of contents might be
wrong. The horrendous link:

http://www.sciencedirect.com/scienc...serid=10&md5=2c8bc9013d6025d08b7d24d3207dfe18

appears to be for the journal itself. The paper does not appear
there, either.

If 2k elements contain a majority element (= an element that occurs at
least k+1 times), then you can find it with 3k-2 element comparisons
(and some small overhead). The basic idea in their algorithm is that
whenever you find two distinct elements, then you can destroy both without
changing the majority element among the remaining elements. This yields:

Run once through the array, and maintain a majority-candidate and a counter.
The majority-candidate is initialized as the first element, and
the counter (counting how often you have seen the candidate) is
initialized at 1.
If you hit the current candidate again, increment the counter.
If you hit another element, decrement the counter.
If the counter becomes 0, drop the current candidate and start from
scratch with the remaining part of the array.

Fischer and Salzberg also show that this bound 3k-2 is best possible in
the worst case (and that's the main part of their paper).

I can find a 1982 technical report from Yale:

FINDING A MAJORITY AMONG N VOTES Solution to Problem 81-5 (Journal
of Algorithms, June 1981)

by the same two authors. It describes what seems to me to be a
different algorithm and for which the 3k - 2 bound is proved. I
suspect the algorithm has been modified since the technical report but
I can't find a good reference that matches the algorithm you describe.

<snip>

 

[mjc]

I was also given this problem as part of an interview quiz. My answer
was: Use one of the known O(n) median algorithms and use that value.
This was accepted by the interviewer (but I didn't get the job).