sort it in quicksort (descending order)...den take arr[1] -->second largest


On Sat, Sep 10, 2011 at 8:34 AM, Akhilesh Vedhera <[email protected]>wrote:

> Then the complexity will be nlogn not n.... and if it is the worst case
> then it would be O(n^2)...
>
> On Sat, Sep 10, 2011 at 8:58 PM, abhinav gupta 
> <[email protected]>wrote:
>
>> Oops ..no u hav to quicksort it.
>>
>>
>> On Sat, Sep 10, 2011 at 8:24 AM, Dave <[email protected]> wrote:
>>
>>> @Abhinav: Does it work correctly on {1, 3, 2}, or, for that matter, on
>>> any array where the second largest comes after the largest?
>>>
>>> Dave
>>>
>>> On Sep 10, 10:16 am, abhinav gupta <[email protected]> wrote:
>>> > temp2 is second largest element.
>>> >
>>> > On Sat, Sep 10, 2011 at 8:00 AM, abhinav gupta <
>>> [email protected]>wrote:
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > > I can solve this problem in O(n)
>>> > > i=0;
>>> > > temp1=arr[0];
>>> >
>>> > > while(i != len)
>>> > > {
>>> > > if(arr[i] > temp1)
>>> > > {
>>> > > temp2=temp1;
>>> > > temp1=arr[i]
>>> > > }
>>> > > i++;
>>> > > }
>>> >
>>> > > On Sat, Sep 10, 2011 at 7:42 AM, Dave <[email protected]>
>>> wrote:
>>> >
>>> > >> @Replying to my own posting: remove the words "one of the numbers
>>> that
>>> > >> lost to", so that the explanation reads
>>> >
>>> > >> The question should be "How can we find the second largest element
>>> in
>>> > >> an array in n + ceiling(log_2(n)) - 2 comparisons?" The answer is to
>>> > >> use a tournament to select the largest number. The second largest
>>> > >> number will have lost to the largest. It takes n - 1 comparisons to
>>> > >> determine the largest number. There are ceiling(log_2(n)) numbers
>>> that
>>> > >> have lost to the largest, and it takes ceiling(log_2(n)) - 1
>>> > >> comparisons to find the largest of them.
>>> >
>>> > >> Dave
>>> >
>>> > >> On Sep 10, 9:28 am, Dave <[email protected]> wrote:
>>> > >> > @Praveen: The question should be "How can we find the second
>>> largest
>>> > >> > element in an array in n + ceiling(log_2(n)) - 2 comparisons?" The
>>> > >> > answer is to use a tournament to select the largest number. The
>>> second
>>> > >> > largest number will have lost to one of the numbers that lost to
>>> the
>>> > >> > largest. It takes n - 1 comparisons to determine the largest
>>> number.
>>> > >> > There are ceiling(log_2(n)) numbers that have lost to the maximum,
>>> and
>>> > >> > it takes ceiling(log_2(n)) - 1 comparisons to find the largest of
>>> > >> > them.
>>> >
>>> > >> > Dave
>>> >
>>> > >> > On Sep 10, 9:18 am, praveen raj <[email protected]> wrote:
>>> >
>>> > >> > > How can we find second largest element in an array... in O(n
>>> > >> > > +logn-2)... give me proof.....- Hide quoted text -
>>> >
>>> > >> > - Show quoted text -
>>> >
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>
>
>
> --
> Akhilesh
> NSIT-COE
>
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