Let n = 300,
1. Compare each pair, you'll get 150 numbers which are minimum in their respective pairs
2. Repeat above step now on remaining 150 numbers
3. Finally you'll get minimum number in n-1 comparisons. i.e 299 comparisons
4. After first step we got 150 numbers which were minimum in their respective pairs. At the same time we got 150 numbers that were maximum in their respective pairs. Our maximum number can only belong to this remaining 150 numbers
5. So now applying same algorithm for 150 numbers to find the maximum number
6. This will require n/2 - 1 comparisons i.e 149 comparisons
7. In total, No. Of comparisons to find maximum and minimum with best case complexity is
((n - 1) + (n/2 - 1)) = 3n/2 - 2 = 448
