You are given an array of 1024 elements, minimum number of comparisons…

Concept: Finding the maximum of n elements via a knockout (tournament) arrangement always takes exactly n − 1 comparisons, since each comparison eliminates exactly one candidate and n − 1 eliminations must happen to leave a single winner. Because the second-largest element is bigger than every other element except the largest, the only element it can ever lose a comparison to is the actual largest, so the second-largest must be one of the elements the eventual champion played against directly during the tournament — never someone deeper down who was eliminated earlier by a third element.

1. Comparisons to find the largest among n = 1024 elements = n − 1 = 1024 − 1 = 1023.

2. In a balanced knockout bracket the champion plays exactly log2n rounds, so it faced at most log21024 = 10 opponents directly over the whole tournament.

3. The second-largest must be among those (at most) 10 direct opponents of the champion, so finding the best of that group needs 10 − 1 = 9 comparisons.

4. Total minimum comparisons = 1023 + 9 = 1032.

Cross-check: this matches the standard closed-form result for this problem — the minimum comparisons to find both the largest and second-largest of n elements by tournament method is n + ⌈log2n⌉ − 2. Substituting n = 1024 gives 1024 + 10 − 2 = 1032, confirming the same answer independently.