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

The following solution would take 2(N-1) comparisons:

arr  #array with 'n' elements
first=arr[0]
second=-999999  #large negative no
i=1
while i is less than length(arr):
    if arr[i] greater than first:
        second=first
        first=arr[i]
    else:
        if arr[i] is greater than second and arr[i] less than first:
            second=arr[i]
    i=i+1
print second