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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