Exercise 1: Brute Force vs. Hash Set at n=200, No Duplicates — Possible Solution
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GIVEN
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n = 200 elements, no duplicates present.

STEP 1: BRUTE-FORCE COMPARISONS
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Per Step 2's own formula, n(n-1)/2:

  200 x 199 / 2 = 39,800 / 2 = 19,900 comparisons

Since there are no duplicates at all, this is also the WORST case
(Step 3) - every single comparison must run, since no early match is
ever found to short-circuit the search.

STEP 2: HASH-SET OPERATIONS
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Per Step 7's own approach, exactly one operation (a lookup-and-insert)
per element:

  200 operations

STEP 3: THE RATIO
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19,900 / 200 = 99.5

At n=200, brute force requires 99.5 times more operations than the
hash-set approach to reach the identical (correct) conclusion that no
duplicates exist.

WHY THIS WORKS AS AN ANSWER
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Both operation counts are computed directly from this chapter's own
stated formulas for each approach, the no-duplicates condition is
explicitly connected to Step 3's own worst-case definition for brute
force, and the ratio is computed explicitly rather than left as two
separate, uncompared numbers.