Answer Posted / om

There are two good method to solve this problem; first one
takes 2n-3 comparison and second one takes n+lg(n)-2
comparison only.

1) Run bubble sort loop only for 2 times ..the first time it
will do n-1 comparison and put the largest element at the
end of the array.... then by running second time it will do
n-2 comparison and put the second largest element at the
second last position in the array... so n-1 + n-2 =2n-3
comparison. cool naa... I didn't sort the whole array and
found the second largest element.

2)http://www.seeingwithc.org/topic3html.html
check this site...

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