Solution

To find the second smallest element, first find the smallest element using tournament method.

Finding the smallest element requires:

$n-1$ comparisons

The smallest element can lose comparisons against at most:

$\lceil \log n\rceil$ elements

Now, to find the second smallest among these elements, comparisons required are:

$\lceil \log n\rceil-1$

Total comparisons:

$(n-1)+(\lceil \log n\rceil-1)$

$=n+\lceil \log n\rceil-2$