Find the 2nd largest element in an array with minimum number of comparisons

Question

For an array of size N, what is the number of comparisons required?

Answer 1

Assuming space is irrelevant, this is the smallest I could get it. It requires 2*n comparisons in worst case, and n comparisons in best case:

arr = [ 0, 12, 13, 4, 5, 32, 8 ]
max = [ -1, -1 ]

for i in range(len(arr)):
     if( arr[i] > max[0] ):
        max.insert(0,arr[i])
     elif( arr[i] > max[1] ):
        max.insert(1,arr[i])

print max[1]