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3 * n/2 - 2 comparisons, tournament method
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| x = [1, 5, 3, 9, 10, 6, 0, 11, 21, -10, 5] | |
| small, large = x[0], x[0] | |
| for i in range(1, len(x) - 1): | |
| if x[i] < x[i + 1]: | |
| small = min(x[i], small) | |
| large = max(x[i + 1], large) | |
| print(small, large) | |
| # 3 * n/2 - 2 comparisons, tournament method | |
| """ | |
| If odd, set both to same value like above. | |
| If even, perform one manual comparison to determine the min and max at the start. | |
| if n is odd, we do 3[n/2] comparisons. | |
| if n is even, we perform 1 initial comparison followed by 3(n - 2)/2 for a total of 3n(2 - 2). | |
| The total compairsons at most is 3[n/2] | |
| """ |
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