@all: it will take O(3n) for finding max n min and if and only if all
elements in an array are distinct .
On Mon, Aug 15, 2011 at 1:30 AM, siddharth srivastava
<akssps...@gmail.com>wrote:
>
> 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 <sinv...@gmail.com> 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 <aditi.garg.6...@gmail.com>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 <
>>> aditya.kumar130...@gmail.com> 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 <
>>>> aditya.kumar130...@gmail.com> wrote:
>>>>
>>>>> yes
>>>>>
>>>>>
>>>>> On Mon, Aug 15, 2011 at 12:58 AM, aditi garg <
>>>>> aditi.garg.6...@gmail.com> 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 <sinv...@gmail.com> wrote:
>>>>>>
>>>>>>> no it is 3*n only........ read it again
>>>>>>>
>>>>>>>
>>>>>>> On Mon, Aug 15, 2011 at 12:45 AM, Amir Aavani <amir.aav...@gmail.com
>>>>>>> > 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<
>>>>>>>>> karthikeyan...@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>> sorry O(n^2) s the time complexity
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On 14 August 2011 23:56, shady<sinv...@gmail.com> 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<
>>>>>>>>>>> karthikeyan...@gmail.com> 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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>>>>>
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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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