Use K-Median theorem with k=N
On Mon, Nov 7, 2011 at 11:33 PM, Gene <gene.ress...@gmail.com> wrote:
> Can you explain how a heap helps get the answer? Just to put the
> elements in a heap requires O ( N^2 log (N) ) time.
>
> On Nov 7, 4:12 pm, vikas <vikas.rastogi2...@gmail.com> wrote:
> > I think the best we can have is nlogn solution with the heap approach.
> >
> > On Nov 6, 10:27 pm, Dave <dave_and_da...@juno.com> wrote:
> >
> >
> >
> > > @Mohit: Here is a counterexample:
> >
> > > 10 11 52 53 54
> > > 20 21 112 113 114
> > > 30 31 122 123 124
> > > 40 41 132 133 134
> > > 50 91 142 143 144
> >
> > > The median is 91, and it is not on the anti-diagonal.
> >
> > > Dave
> >
> > > On Nov 6, 3:11 am, mohit verma <mohit89m...@gmail.com> wrote:
> >
> > > > @Gene
> > > > As i said in my earlier post right to left diagonal partitions the
> martix
> > > > into 2 equal number of elements. So now the median must be in this
> > > > diagonal. Now our focus is on finding median of this diagonal only.
> > > > I think this works fine. Can u give some test case for which it
> fails?
> >
> > > > On Sun, Nov 6, 2011 at 3:02 AM, Gene <gene.ress...@gmail.com> wrote:
> > > > > Unfortunately this isn't true. See the example I gave earlier:
> >
> > > > > 1 2 3
> > > > > 2 4 5
> > > > > 3 4 6
> >
> > > > > Thje median is 3.
> >
> > > > > 1 2 2 3 >3< 4 4 5 6
> >
> > > > > Niether one of the 3's lies on the diagonal.
> >
> > > > > When you pick any element P on the diagonal, all you know is that
> > > > > anything to the right and downward is no less than P and
> everything to
> > > > > the left and upward is no greater. This leaves the upper right and
> > > > > lower left rectangles of the matrix unrelated to P.
> >
> > > > > On Nov 5, 3:51 am, ankit agarwal <ankit.agarwal.n...@gmail.com>
> wrote:
> > > > > > Hi,
> >
> > > > > > I think the median will always lie on the diagonal
> > > > > > a[n][1] ---- a[1][n]
> > > > > > because the elements on the LHS making the upper triangle will
> > > > > > always be less than or equal to the elements on the diagonal
> > > > > > and the RHS, elements in the lower triangle will be greater than
> or
> > > > > > equal to them.
> >
> > > > > > so sort the diagonal and find the middle element, that will be
> the
> > > > > > median.
> >
> > > > > > Thanks
> > > > > > Ankit Agarwal
> >
> > > > > > On Nov 5, 1:29 am, Gene <gene.ress...@gmail.com> wrote:
> >
> > > > > > > Here's an idea. Say we pick any element P in the 2D array A
> and use
> > > > > > > it to fill in an N element array X as follows.
> >
> > > > > > > j = N;
> > > > > > > for i = 1 to N do
> > > > > > > while A(i, j) > P do
> > > > > > > j = j - 1;
> > > > > > > end;
> > > > > > > X(i) = j;
> > > > > > > end;
> >
> > > > > > > This algorithm needs O(N) time.
> >
> > > > > > > The elements of X split each row with respect to P. That is,
> for each
> > > > > > > i = 1 to N,
> >
> > > > > > > A(i, j) <= P if 0 < j <= X(i), A(i,j) > P if X(i) < j <=
> N.
> >
> > > > > > > Now the strategy is to create two length N arrays a =
> [0,0,...0]; and
> > > > > > > b = [N,N,...]. We'll maintain the invariant that a[i] < Median
> <= b[i]
> > > > > > > for some i. I.e, they "bracket" the median.
> >
> > > > > > > We define functions L(a) = sum_i( a(i) ) and R(b) = sum_i( N -
> > > > > > > b(i) ). These tell us how many elements there are left and
> right of
> > > > > > > the bracket.
> >
> > > > > > > Now reduce the bracket as in binary search: Guess a value P,
> compute
> > > > > > > X. If L(X) >= R(X), set b = X else set a = X.
> >
> > > > > > > Keep guessing new P values in a way that ensures we reduce the
> number
> > > > > > > of elements between a and b by some fixed fraction. If we can
> do
> > > > > > > that, we'll get to 1 element in O(N log N) time.
> >
> > > > > > > The remaining problem is picking good P's. Certainly the first
> time is
> > > > > > > easy. Just take A(N/2, N/2). This has approximately (at least)
> N^2/4
> > > > > > > elements larger than it and N^2/4 smaller due to the sorted
> rows and
> > > > > > > columns. This is what we need to get O(N log N) performance.
> >
> > > > > > > But after the first split, things get trickier. The area
> between a and
> > > > > > > b takes on the shape of a slash / /, so you can't just pick a
> P that
> > > > > > > moves a and b together by a fixed fraction of remaining
> elements.
> >
> > > > > > > Not to worry! You can quickly look up the (at most) N row
> medians in
> > > > > > > the bracket, i.e.
> >
> > > > > > > { A(i, (a[i] + b[i] + 1) / 2) | a[i]<b[i] , i = 1 to N }
> >
> > > > > > > and use the well known O(N) median selection algorithm to get
> a median
> > > > > > > of this. This has the quality we want of being somewhere
> roughly in
> > > > > > > the middle half of the remaining elements. The logic is the
> same as
> > > > > > > the selection algorithm itself, but in our case the rows are
> pre-
> > > > > > > sorted.
> >
> > > > > > > In all, each partitioning step requires O(N), and a fixed
> fraction
> > > > > > > (about 1/2) of the elements will be eliminated from the
> bracket with
> > > > > > > each step. Thus O(log n) steps will be needed to bring the
> bracket to
> > > > > > > size 1 for an overall cost of O(N log N).
> >
> > > > > > > I don't doubt that there's a simpler way, but this one seems
> to work.
> > > > > > > Anyone see problems?
> >
> > > > > > > On Nov 3, 3:41 pm, sravanreddy001 <sravanreddy...@gmail.com>
> wrote:
> >
> > > > > > > > any better solution than O(N^2) in worst case?
> > > > > > > > How do we take advantage of sorting and find in O(N lg N)-
> Hide
> > > > > quoted text -
> >
> > > > > > - Show quoted text -
> >
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> > > > --
> > > > Mohit- Hide quoted text -
> >
> > - Show quoted text -
>
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