Pairs with a given sum: Θ(n²) versus Θ(n)

Problem

Given an array of n integers and a target, count the pairs of positions i < j with a[i] + a[j] == target. Solve the problem twice and count the work each method does — do not measure time.

Brute force: two nested loops over all pairs i < j. Count one comparison for every pair examined.
Hash map: scan the array once, keeping a Map<Long, Long> from each value seen so far to how often it has been seen. For each element x, add the number of times target - x has been seen, then record x. Count one lookup per element.

Input: n, then n integers, then target (values fit in an int, but sums may not — use long). Output exactly:

Brute force: <pairs> pairs, <comparisons> comparisons
Hash map: <pairs> pairs, <lookups> lookups

For the input

5
1 5 7 -1 5
6

the output is

Brute force: 3 pairs, 10 comparisons
Hash map: 3 pairs, 5 lookups

(the pairs are 1 + 5, 1 + 5 and 7 + (−1)). Both methods must always report the same number of pairs.

Your solution

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