but how does it use the fact that the matrix was initially sorted row-wise
and column-wise? that's far from efficient.
On Tue, Jan 8, 2013 at 11:44 PM, Amit Basak <abas...@gmail.com> wrote:
> Take a one dimensional array as the multiplication of both the dimensions
> of the two dimensional array and copy the two dimensional array elements in
> the one dimensional array.
> After that use an efficient sorting (e.g. quick sort) on this one
> dimensional array
>
> Regards,
> - Amit
> -------------------
> Sent from Nexus 4
> On Jan 8, 2013 11:30 PM, "Ravi Ranjan" <ravi.cool2...@gmail.com> wrote:
>
>> You have a two dimensional array of size m*n. The
>> elements in the rows are sorted and every row has
>> unique elements means( in a row no two elements are same) but
>> the elements can repeat across the rows.
>> For example:
>> you have following 2-D array:
>> 2 5 7 9 14 16
>> 3 6 8 10 15 21
>> 4 7 9 15 22 35
>> 7 8 9 22 40 58
>> You are supposed to write an efficient function which
>> will take upper 2-D array as input and will return a
>> one-dimensional array with following properties:
>> a) the 1-D array must contain all the elements of above
>> 2-D array.
>> b) the 1-D array should not have any repeated elements.
>> c) all the elements should be in sorted order.
>> For example:
>> for the above 2-D array, the output should be:
>> A [ ] = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 14, 15, 16, 21, 22, 35,
>> 40, 58 }
>>
>> --
>>
>>
>>
> --
>
>
>
--
- Saurabh Paliwal
B-Tech. Comp. Science and Engg.
IIT ROORKEE
--
