How do you partition an array into 2 parts such that the two parts have equal average? Each partition may contain elements that are non-contiguous in the array. The only algorithm I can think of is exponential can we do better?
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2Be honest, is this a homework question? – Brian Driscoll Jan 18 '12 at 17:59
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4What have you tried? Did it work? Have you got some test cases with example input and output? – Greg Hewgill Jan 18 '12 at 18:00
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6this sounds more like an interview question, not an easy one at that – BrokenGlass Jan 18 '12 at 18:05
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The only algorithm I can think of is exponential can we do better? – CommonMan Jan 18 '12 at 18:11
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This question is especially skimpy on details. Let's start with the basics, for example *is it guaranteed* that there is such a partition? If it's a verbatim interview question and you do not immediately ask for more information I 'd say you failed the question. – Jon Jan 18 '12 at 18:17
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1What *is* the algorithm that you can think of? – YXD Jan 18 '12 at 18:22
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1brute force testing all subsets would calculate a solution if there is one - but this is O(2**N) – BrokenGlass Jan 18 '12 at 18:25
1 Answers
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You can reduce this problem to the sum-subset problem - also cached here. Here's the idea.
Let A be the array. Compute S = A[0] + ... + A[N-1], where N is the length of A. For k from 1 to N-1, let T_k = S * k / N. If T_k is an integer, then find a subset of A of size k that sums to T_k. If you can do this, then you're done. If you cannot do this for any k, then no such partitioning exists.
Here's the math behind this approach. Suppose there is a partitioning of A such that the two parts have the same average, says X of size x and Y of size y are the partitions, where x+y = N. Then you must have
sum(X)/x = sum(Y)/y = (sum(A)-sum(X)) / (N-x)
so a bit of algebra gives
sum(X) = sum(A) * x / N
Since the array contains integers, the left hand side is an integer, so the right hand side must be as well. This motivates the constraint that T_k = S * k / N must be an integer. The only remaining part is to realize T_k as the sum of a subset of size k.
PengOne
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after a bit of thinking i still don't see how your proof reduces subset-sum to OP's problem. – soulcheck Jan 18 '12 at 19:45
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i mean, you essencially proved that having answer to subset sum one can answer OP's problem, not the other way around. – soulcheck Jan 18 '12 at 19:57
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@soulcheck I'm still pondering whether they are truly equivalent. I believe the answer comes down to [this question I just asked](http://stackoverflow.com/questions/8916539/sum-subset-with-a-fixed-subset-size). – PengOne Jan 18 '12 at 19:59
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@PengOne I am confused whether we can use sum-subset problem to find the set of size k which sums to T_k. What if there are multiple sets of different size that sums to T_k and one such set is of size k. Is there a way we can tweak the sum_subset problem to find if there exists a set of size k? – NPE Sep 06 '12 at 03:26
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@PengOne Is there a solution approach for this question? I wrote a code following this approach but its churning the wrong output. My code: https://ideone.com/HadH3V – user248884 Sep 07 '18 at 14:07