Sorry , I've tried but BS will not work here .
On Tue, Jul 24, 2012 at 9:17 PM, algo bard <algo.b...@gmail.com> wrote:
> @Shobhit: Can you give me a few hints on implementing a BS on the 2D?
> @neelpulse: That's what I said. A 2D array *might* be a probable
> candidate. In your example, the first 2d satisfies the criteria...so we
> check it -- Not found -- Reject -- Move on to next probable candidate.
>
> On Sat, Jul 21, 2012 at 5:14 PM, neelpulse(Jadavpur University) <
> neelpu...@gmail.com> wrote:
>
>> May be I am missing a few details. But Consider this 3D array:
>> {
>> {
>> {1,2},
>> {7,8} // First 2D array
>> },
>> {
>> {3,4},
>> {9,10}
>> }
>> }
>> If you search for 3 then your search in first step will give first 2D
>> which actually does not contain 3. As per my interpretation of the problem,
>> my array is holding the preconditions.
>>
>> On Friday, 20 July 2012 16:25:49 UTC+5:30, algo bard wrote:
>>>
>>> Compare the element with the first([0][0]) and the last
>>> element([n-1][n-1]) of each 2D array to pin down the 2D array it *might* be
>>> present in.
>>> After that you can follow this approach : http://www.geeksforgeeks.org/
>>> **archives/11337 <http://www.geeksforgeeks.org/archives/11337>
>>>
>>> If it's not present in that 2D, move on and search for the next target
>>> 2D.
>>>
>>> The Probable 2D target set will be given by :
>>> arr[i][0][0]<=element<=arr[i][**n-1][n-1].
>>> Reject the 2Ds which don't follow this condition.
>>>
>>> TC: O(n^2)
>>>
>>> Though, I think an O(n) approach must exist for this problem.
>>>
>>> On Fri, Jul 20, 2012 at 11:24 AM, Sakshi Agrawal <
>>> sweetsaksh...@gmail.com> wrote:
>>>
>>>> How will you search an element in sorted 3D Array ? ( Sorted in all
>>>> the 3 directions )
>>>>
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