I think if x < a[n/2][n/2] than we need to search in 1st quadrant otherwise
in others.

On Wed, Jan 5, 2011 at 7:20 PM, sourabh jakhar <sourabhjak...@gmail.com>wrote:

> aditya solution is correct
> it is a standard question of young tabuleau
> it is complexity is log(n)
>
>
> On Wed, Jan 5, 2011 at 6:52 PM, ADITYA KUMAR <aditya...@gmail.com> wrote:
>
>> @MAC
>> ur solution is wrong
>>
>> 1 9 24
>> 2 12 33
>> 3 16 49
>>
>> search for 3
>>
>>
>> O(logn) solution
>> let the matrix be A[][] and number to be searched is x
>> divide the array from middle in 4 parts lets say it four quadrants
>> now check if x>A[n/2][n/2] search in bottom right quadrant
>> if x<A[n/2][n/2] search in other 3 quadrants
>>
>> On Sat, Dec 25, 2010 at 8:25 AM, yq Zhang <zhangyunq...@gmail.com> wrote:
>>
>>> Suppose you have a matrix n*m. each column and row of the matrix is
>>> already sorted. For example:
>>>
>>> 1,2,3
>>> 2,3,4
>>> 4,5,6
>>>
>>> All 3 rows and 3 columns of above matrix are sorted. How to find a
>>> specific number in the matrix?
>>> The trivial O(nlogm) solution is to use binary search for all rows. I
>>> am looking for better solution.
>>>
>>> Thanks
>>>
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>>
>>
>> --
>> Regards
>> Aditya Kumar
>> B-tech 3rd year
>> Computer Science & Engg.
>> MNNIT, Allahabad.
>>
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>
>
>
> --
> SOURABH JAKHAR,(CSE)(3 year)
> ROOM NO 167 ,
> TILAK,HOSTEL
> 'MNNIT ALLAHABAD
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-- 
Cheers
Naveen Kumar

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