yes could be done by taking struct for each element
struct st{int a; //value of elementint i;
//index of beaten arrayint beaten[90]; //contains all beaten
elements beaten by this element }ele[90];
here is the modified code :- http://ideone.com/FqnSq
On Mon, Sep 3, 2012 at 8:10 PM, sangeeta goyal <[email protected]>wrote:
> @Darpan can you implement it without using the multimap and with the same
> (divide and conquar) approach.
>
>
> On Mon, Sep 3, 2012 at 6:25 PM, Darpan Baweja <[email protected]>wrote:
>
>> hope this might helps
>> code:- http://ideone.com/mtHem
>> used divide and conquer approach
>> and stored all the beaten elements in the multimap with key is the
>> element which has beaten them
>> with key as winner return maximum element stored in multimap
>>
>> On Mon, Sep 3, 2012 at 3:31 PM, sangeeta goyal
>> <[email protected]>wrote:
>>
>>> @bharat is it tournament method??
>>>
>>>
>>> On Mon, Sep 3, 2012 at 2:34 PM, bharat b
>>> <[email protected]>wrote:
>>>
>>>> Construct a max-heap --> O(n)..
>>>> call delete() 2 times .. --> O(logn)..
>>>> ===> O(n) time..
>>>>
>>>>
>>>> On Fri, Aug 31, 2012 at 1:46 AM, Don <[email protected]> wrote:
>>>>
>>>>> While the list length is more than one
>>>>> Take 2 elements from the head
>>>>> Select the larger of the two
>>>>> If the smaller is greater than the largest beaten by the larger
>>>>> Then set the largest beaten by the larger to the value of
>>>>> the smaller
>>>>> Add the larger to the tail of the list
>>>>>
>>>>> When this completes, you'll have one element containing the largest
>>>>> and second largest values.
>>>>>
>>>>> typedef struct
>>>>> {
>>>>> unsigned int value;
>>>>> unsigned int largestBeaten;
>>>>> } element;
>>>>>
>>>>> unsigned int secondLargest(queue<element> elements)
>>>>> {
>>>>> while(elements.length() > 1)
>>>>> {
>>>>> element A = elements.dequeue();
>>>>> element B = elements.dequeue();
>>>>> if (A.value < B.value) swap(A,B);
>>>>> if (A.largestBeaten < B.value) A.largestBeaten = B.value;
>>>>> elements.enqueue(A);
>>>>> }
>>>>> return queue.head().largestBeaten;
>>>>> }
>>>>>
>>>>> On Aug 30, 12:53 pm, sangeeta goyal <[email protected]> wrote:
>>>>> > @Don can you give the algorithm for the same??
>>>>> > how would you implement it??
>>>>> >
>>>>> >
>>>>> >
>>>>> >
>>>>> >
>>>>> >
>>>>> >
>>>>> > On Thu, Aug 30, 2012 at 10:03 PM, Don <[email protected]> wrote:
>>>>> > > The second largest element is the largest element beaten by the
>>>>> > > winner.
>>>>> > > So if you implement a tournament in which each element keeps track
>>>>> of
>>>>> > > the largest element it has beaten, you'll get the second largest
>>>>> > > naturally.
>>>>> > > Don
>>>>> >
>>>>> > > On Aug 29, 9:15 am, Sangeeta <[email protected]> wrote:
>>>>> > > > give the algo or program to find second largest element in a
>>>>> list using
>>>>> > > > tournament method
>>>>> >
>>>>> > > --
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>>
>> --
>> *DARPAN BAWEJA*
>> *Final year, I.T*
>> *NIT Allahabad*
>>
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