Find Second largest number in array at most n+log₂(n)−2 comparisons [closed]

Question

You are given as input an unsorted array of n distinct numbers, where n is a power of 2. Give an algorithm that identifies the second-largest number in the array, and that uses at most n+log₂(n)−2 comparisons.

Answer 1

1. Start with comparing elements of the n element array in odd and even positions and determining largest element of each pair. This step requires n/2 comparisons. Now you've got only n/2 elements. Continue pairwise comparisons to get n/4, n/8, ... elements. Stop when the largest element is found. This step requires a total of n/2 + n/4 + n/8 + ... + 1 = n-1 comparisons.
2. During previous step, the largest element was immediately compared with log₂(n) other elements. You can determine the largest of these elements in log₂(n)-1 comparisons. That would be the second-largest number in the array.

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Example: array of 8 numbers [10,9,5,4,11,100,120,110].

Comparisons on level 1: [10,9] ->10 [5,4]-> 5, [11,100]->100 , [120,110]-->120.

Comparisons on level 2: [10,5] ->10 [100,120]->120.

Comparisons on level 3: [10,120]->120.

Maximum is 120. It was immediately compared with: 10 (on level 3), 100 (on level 2), 110 (on level 1).

Step 2 should find the maximum of 10, 100, and 110. Which is 110. That's the second largest element.