Constructing a max-heap is O(n*log n)
However, the problem asked for a solution using tournament method.
If you ignore that requirement, an O(n) solution is trivial, and
doesn't require the additional storage of a heap:
int first = max(A[0], A[1]);
int second = min(A[0], A[1]);
for(i = 2; i < N; ++i)
{
if (A[i] >= first)
{
second = first;
first = A[i];
}
else if (A[i] > second)
second = A[i];
}
// First and second now contain 1st and 2nd largest values in A
Don
On Sep 3, 5:04 am, bharat b <bagana.bharatku...@gmail.com> wrote:
> Construct a max-heap --> O(n)..
> call delete() 2 times .. --> O(logn)..
> ===> O(n) time..
>
>
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