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 - > > > > -- > > > You received this message because you are subscribed to the Google Groups > > > "Algorithm Geeks" group. > > > To post to this group, send email to algogeeks@googlegroups.com. > > > To unsubscribe from this group, send email to > > > algogeeks+unsubscr...@googlegroups.com. > > > For more options, visit this group at > > >http://groups.google.com/group/algogeeks?hl=en. > > > -- > > Mohit -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. 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