Well, friend, we're both wrong.
The algorithm will find 6 just fine. It will choose 3 as the middle
element. Since 6 is bigger, it will throw away the subarray
1 2
2 3
and check the other 3 subarrays. When it checks
6 7
7 8
It will find the 6 on the first try. I just verified this by running
the code.
Second, I should have solved the recurrence. You're right that it's ~
n^1.6 . Throwing away a quarter of the _elements_ isn't good enough
because that number is n^2.
The algorithm is only sublinear in the number of elements, which of
course is worse than the standard algorithm that starts at the lower
left corner and steps along a jagged path of max length ~2n.
But I should have seen that you can split the L-shaped remaining piece
of each subarray into only TWO pieces rather than 3. So the recursive
calls should have been:
return (x < a[mi][mj]) ?
search(a, x, i0, mi - 1, j0, j1, i, j) ||
search(a, x, mi, i1, j0, mj - 1, i, j)
:
search(a, x, i0, mi, mj + 1, j1, i, j) ||
search(a, x, mi + 1, i1, j0, j1, i, j);
With this change the recurrence is more complicated, but it should be
faster than n^1.6 . I'm not going to try it tonight...
On Sep 27, 8:19 am, Nikhil Jindal <[email protected]> wrote:
> @saurabh.nsit:
>
> Consider the following array:
>
> 1 2 6 7
> 2 3 7 8
> 4 5 8 9
> 5 7 9 10
>
> And the item to be searched is 6. As I understand it, using your approach
> you will search 6 in only the second and third row, which will not give the
> correct solution.
> Hope this clears a few doubts.
>
> @Gene:
> Analysing the complexity of ur algo:
>
> T(n) = 3*T(n/2) + O(1)
>
> which is n^(log_2(3)) = n^1.6.
>
> Cheers
> Nikhil Jindal
>
> On Sun, Sep 26, 2010 at 11:14 PM, saurabh singh <[email protected]>wrote:
>
>
>
>
>
> > As you mentioned ultimately element to be searched should either be in row
> > 'i' (ahead of [i,i] element) or in row i+1 (before [i+1,i+1] element). Since
> > each row contain numbers in sorted order so u can do binary search on these
> > two rows and ultimately the complexity will be O(logn) only
>
> > On Sun, Sep 26, 2010 at 7:34 PM, Nikhil Jindal
> > <[email protected]>wrote:
>
> >> On Tue, Sep 21, 2010 at 6:05 PM, saurabh singh
> >> <[email protected]>wrote:
>
> >>> solution 1:
> >>> use concept of quad-tree and do binary search in that tree
>
> >>> solution 2:
> >>> do binary search on major diagonal. ultimately u will narrow down to
> >>> search for element in 2 rows. in these two rows again do binary search.
>
> >> How do you narrow down to two rows? Please explain.
> >> By searching on the diagonal, you get two elements such that one is lesser
> >> than the number being searched for and the next is greater. let them be
> >> i,i,
> >> and i+1,i+1.
>
> >> So you remove the array from 0,0 to i,i and from i+1,i+1 to n-1,n-1. But
> >> the number could be anywhere in the rest of the array
>
> >>> any solution will lead you to O(log(n)) time
>
> >>> On Tue, Sep 21, 2010 at 5:10 PM, jagadish <[email protected]>wrote:
>
> >>>> Hi all,
> >>>> Given a 2d array which is sorted row wise and column wise as well,
> >>>> find a specific element in it in LESS THAN O(n).
> >>>> PS: an O(n) solution would involve skipping a column or a row each
> >>>> time from the search and moving accordingly.
> >>>> Solution less than O(n) is desirable!
>
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> >>> --
> >>> Thanks & Regards,
> >>> Saurabh
>
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