Consider n=8 elements in an array {1,4,5,8,3,2,7,9}
Lets make a tournament bracket for them, where at each stage the winner is the minimum element between the two.
As you can see, number of comparisons being done = n-1 = 7
Similarly, to find the maximum element you again will need n-1 comparisons!
So total no of comparisons to find min and max=2(n-1)
Hmm...there is one optimisation to it !!
The last level in the tree is making n/2 comparisons(4 in this case) and these are being repeated while finding the minimum and maximum! So doing the last level comparisons only once, we do n/2 comparisons less
Hence 2(n-1) - n/2 = 2n-2-n/2 = 3n/2-2
Source:https://www.quora.com/How-does-the-tournament-method-for-finding-the-maximum-and-minimum-element-in-an-array-consist-of-3n-2-2-comparisons