One approach could be:

I think the min and max elements can be found in n*n*(3n/2)
keep a flag(or array of flag--bitvector ??) for the rows in which min or max
was found
print all other rows for which flag is not set

another n^3(confirm please) algo could be:
sort each row individually
compare first col for min, last col for max
print all which do not have min or max



On 15 August 2011 01:20, shady <[email protected]> wrote:

> according to me it would be take 4*n time.... 3 iterations to choose the
> min. and max. from 1st three rows, and n again to print the chosen one
>
>
> On Mon, Aug 15, 2011 at 1:13 AM, aditi garg <[email protected]>wrote:
>
>> some how im not able to get the logic...how will i be able to find max and
>> min of the entire matrix by jst traversing 3 rows??
>>
>> for eg
>>  1 2 3 4 5
>>  8 1 3 6 9
>> 4 6 3 2 10
>> 9 0 5 8 12
>> 18 2 6 7 3
>> fr dis matrix how will u find max and min??
>>
>>
>> On Mon, Aug 15, 2011 at 1:04 AM, aditya kumar <
>> [email protected]> wrote:
>>
>>> just traverse the three rows and get the max and min out of the three
>>> rows . print the row in which their is no max and min .
>>>
>>>
>>> On Mon, Aug 15, 2011 at 1:02 AM, aditya kumar <
>>> [email protected]> wrote:
>>>
>>>> yes
>>>>
>>>>
>>>> On Mon, Aug 15, 2011 at 12:58 AM, aditi garg <[email protected]
>>>> > wrote:
>>>>
>>>>> @shady : does O(3n) include the time required to find the max and min
>>>>> element as well??
>>>>>
>>>>>
>>>>> On Mon, Aug 15, 2011 at 12:50 AM, shady <[email protected]> wrote:
>>>>>
>>>>>> no it is 3*n only........ read it again
>>>>>>
>>>>>>
>>>>>> On Mon, Aug 15, 2011 at 12:45 AM, Amir Aavani 
>>>>>> <[email protected]>wrote:
>>>>>>
>>>>>>>
>>>>>>> On 08/14/2011 11:46 AM, aditya kumar wrote:
>>>>>>>
>>>>>>>> it can be done in O(3n). in worst case one row will have max and
>>>>>>>> anothr row
>>>>>>>> will have min so the third row will be your o/p to print
>>>>>>>>
>>>>>>> Do you mean O(n^3)?
>>>>>>>
>>>>>>> Consider this { O(n^2) }:
>>>>>>>
>>>>>>>  1- Scan the whole matrix and find minimum and maximum entries in the
>>>>>>> matrix. Let Delta be the difference between maximum and minimum.
>>>>>>>  2- For each row, find the minimum and maximum entries in that row.
>>>>>>> If their difference is exactly Delta, then print that row.
>>>>>>>
>>>>>>>
>>>>>>> Amir
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>> On Mon, Aug 15, 2011 at 12:00 AM, Karthikeyan palani<
>>>>>>>> [email protected]>  wrote:
>>>>>>>>
>>>>>>>>  sorry O(n^2) s the time complexity
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On 14 August 2011 23:56, shady<[email protected]>  wrote:
>>>>>>>>>
>>>>>>>>>  how can it be O(n) when there are itself n*n elements..
>>>>>>>>>>
>>>>>>>>>> PS : no sharing of code, else the inevitable
>>>>>>>>>>
>>>>>>>>>> On Sun, Aug 14, 2011 at 11:51 PM, Karthikeyan palani<
>>>>>>>>>> [email protected]>  wrote:
>>>>>>>>>>
>>>>>>>>>>  Given a n x n matrix. .number are randomly placed. .print any one
>>>>>>>>>>> row
>>>>>>>>>>> which doesn’t have min
>>>>>>>>>>> and max elements. Time Complexity : 0(n)
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> if anyone know the code.. pls share!!!
>>>>>>>>>>>
>>>>>>>>>>> --
>>>>>>>>>>> karthikeyankkn
>>>>>>>>>>>
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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Aditi Garg
>>>>> Undergraduate Student
>>>>> Electronics & Communication Divison
>>>>> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
>>>>> Sector 3, Dwarka
>>>>> New Delhi
>>>>>
>>>>>
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>>
>>
>> --
>> Aditi Garg
>> Undergraduate Student
>> Electronics & Communication Divison
>> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
>> Sector 3, Dwarka
>> New Delhi
>>
>>
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-- 
Regards
Siddharth Srivastava

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