Finding largest and $2^\text{nd}$ largest number in a list

Complexity of this algorithm is $n + log_2(n) - 2$

• minimum # of comparisons $= 128 + 7 - 2 = 133$
• The 128 numbers are divided into two sets of 64 each, perform pairwise comparison keep minimum set of numbers
• The 64 numbers are divided into two sets of of 32 .....

• ....... and so on till you reach to 4

• the above steps requires 64 + 32 + 16 +8 +4 + 2 = 126 comparisons to get the maximum, $= n -1$

To get the second maximum number, you need to check only those elements which were compared with the maximum number in the above part. So to find the 2nd maximum you need $log_2(n) - 1$

total = $n + log_2(n) - 2$