I don't think it is O(N*K)
It is O(N^2)
Just to confirm if i have understood correctly
Lets say n and k are given and are global to all subproblems
Subproblem Definition
F[i,j] -> maximum expected sum that we can get by partitioning array from 0
to i into j subarrays. Each subarray satisfies the size bound
[ceiling(n/2k),floor(3n/2k)]
F[i,j] = max over all r in our size bound {F[i-r,j-1] + expected sum of
subarray from i-r+1 to i}
So to calculate every subsequent subproblem we need to check solutions to
O(N/k) previous subproblems. And there are N subproblems, O((1/k)*N^2)
On Mon, Nov 28, 2011 at 6:46 PM, Dumanshu <duman...@gmail.com> wrote:
> Hi
>
> I may not have understood your solution properly. But i think that
> your solution is an implementation of brute force where you are
> dealing with all cases valid in the range n/2k and 3n/2k without any
> optimization with regard to DP. Am i right?
>
> On Nov 28, 2:17 am, sourabh <sourabhd2...@gmail.com> wrote:
> > Consider the example that you have given:
> > [0,0,1,1,0,0,1,1,0,1,1,0] , here n = 12 and k=3
> >
> > Now we need to partition the array into 3 contiguous sub arrays such
> > that :
> > a) The expected sum value is maximum
> > b) and the size of each sub array should be between 2 and 6, both
> > inclusive. In case, this constraint is not satisfied then its not a
> > valid candidate for the solution even if the partition produces the
> > max value.
> >
> > 2 = ceil (n / 2k) = ceil (12/6)
> > 6 = floor (3n / 2k) = floor (36/6)
> > ---------------------------------------------
> > As mentioned above the following equation :
> > F(n,k) = MAX for all r such that ceil(n/2k) <= r <= floor(3n/2k)
> > { (expected value of array elems from A[n] to A[n-r+1]) + F(n-r,
> > k-1) }
> >
> > /**
> > For understanding how partitioning of the array is represented by the
> > above equation:
> >
> > Say there is an array denoted by A[i] and it needs to be divided into
> > 3 contiguous parts, one of the ways to do so would be to take the
> > following steps :
> >
> > Let K(partition no.) be initialized to 3.
> > Let array size N be 12.
> >
> > a) If N is 0, the goto step 'f'
> > b) If K == 1 then call it as partition K and goto step 'e'.
> > c) Lets take X no. of elements from the end of array A of size N and
> > call it partition K.
> > d) Decrement K by 1 and N by X { --K; and N-=X;}
> > d) Goto step 'a'
> > e) Valid partition and End.
> > f) Not a valid partition and End.
> >
> > Now if the above set of steps is run for all values of X such that
> > 2<=X<=6 , then it will generate all possible candidates (partitions)
> > as per the given problem statement. And for all the valid
> > partitions(the ones that will hit step 'e') we need to calculate the
> > expected sum value.
> > **/
> >
> > can be translated into,
> > // I am using 1-based array indexing for better clarity
> > // A[x .. y] means all elements from A[y] to A[x]..
> >
> > F(12, 3) = MAX
> > {
> > ExpVal (A[12 .. 11]) + F(10, 2) ,
> > ExpVal (A[12 .. 10]) + F(9, 2) ,
> > ExpVal (A[12 .. 9]) + F(8, 2) , //
> > this will yield the maximum sum..
> > ExpVal (A[12 .. 8]) + F(7, 2) ,
> > ExpVal (A[12 .. 7]) + F(6, 2)
> > }
> >
> > which is nothing but,
> > F(12, 3) = MAX
> > {
> > 1/2 + F(10, 2) ,
> > 2/3 + F(9, 2) ,
> > 2/4 + F(8, 2) , // this will yield the
> > maximum sum..
> > 3/5 + F(7, 2) ,
> > 4/6 + F(6, 2)
> > }
> >
> > Trace the above equation and you should get it..
> >
> > On Nov 28, 12:57 am, Ankur Garg <ankurga...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Hey Sourabh
> >
> > > Could you please explain the solution in a bit detail perhaps using an
> > > example or so..It wud be really helpful ..Just logic not code
> >
> > > On Mon, Nov 28, 2011 at 1:03 AM, sourabh <sourabhd2...@gmail.com>
> wrote:
> > > > Looks like a dynamic programming problem....
> >
> > > > Say F(n,k) denotes the maximum expected sum value for an array of n
> > > > elements and partition k , then
> >
> > > > F(n,k) = MAX for all r such that ceil(n/2k) <= r <= floor(3n/2k)
> > > > { (expected value of array elems from A[n] to A[n-r+1]) + F(n-r,
> > > > k-1) }
> >
> > > > Base condition:
> > > > 1) F(N, 1) = expected value for array A[n] such that ceil(n/2k) <= N
> > > > <= floor(3n/2k)
> > > > 2) If any of the sub problems where the array size is not between
> > > > ceil(n/2k) and floor(3n/2k) , both inclusive, then its not a valid
> > > > candidate for the final solution. This is can be handled by giving
> > > > initial value to all such combination a value of -1.
> >
> > > > To store that the intermediate computations take an array Max[N][K],
> > > > F(N,K) = Max[N][K]
> >
> > > > On Nov 28, 12:17 am, sourabh <sourabhd2...@gmail.com> wrote:
> > > > > Because in the previous example k = 3.
> >
> > > > > On Nov 27, 10:46 pm, Piyush Grover <piyush4u.iit...@gmail.com>
> wrote:
> >
> > > > > > Optimal split: [0,0][1,1][0,0][1,1][0,1][1,0]
> > > > > > Expected value of optimal split: 0 + 1 + 0 + 1 + 1/2 + 1/2 = 3
> > > > > > why this is not the optimal split???
> >
> > > > > > On Sun, Nov 27, 2011 at 6:58 PM, Ankur Garg <
> ankurga...@gmail.com>
> > > > wrote:
> > > > > > > You have an array with *n* elements. The elements are either 0
> or 1.
> > > > You
> > > > > > > want to *split the array into kcontiguous subarrays*. The size
> of
> > > > each
> > > > > > > subarray can vary between ceil(n/2k) and floor(3n/2k). You can
> > > > assume that
> > > > > > > k << n. After you split the array into k subarrays. One
> element of
> > > > each
> > > > > > > subarray will be randomly selected.
> >
> > > > > > > Devise an algorithm for maximizing the sum of the randomly
> selected
> > > > > > > elements from the k subarrays. Basically means that we will
> want to
> > > > split
> > > > > > > the array in such way such that the sum of all the expected
> values
> > > > for the
> > > > > > > elements selected from each subarray is maximum.
> >
> > > > > > > You can assume that n is a power of 2.
> >
> > > > > > > Example:
> >
> > > > > > > Array: [0,0,1,1,0,0,1,1,0,1,1,0]
> > > > > > > n = 12
> > > > > > > k = 3
> > > > > > > Size of subarrays can be: 2,3,4,5,6
> >
> > > > > > > Possible subarrays [0,0,1] [1,0,0,1] [1,0,1,1,0]
> > > > > > > Expected Value of the sum of the elements randomly selected
> from the
> > > > subarrays: 1/3 + 2/4 + 3/5 = 43/30 ~ 1.4333333
> >
> > > > > > > Optimal split: [0,0,1,1,0,0][1,1][0,1,1,0]
> > > > > > > Expected value of optimal split: 1/3 + 1 + 1/2 = 11/6 ~
> 1.83333333
> >
> > > > > > > Source ->
> > > >
> http://stackoverflow.com/questions/8189334/google-combinatorial-optim...
> >
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