Given two slices of integers, return the values that appear in exactly one of them (in the first but not the second, or the second but not the first), with no duplicates, sorted ascending. This is the symmetric difference of the two sets.
exclusiveItems(a, b)
Input format: a | b. Output the exclusive values, sorted, space-separated
(empty line if none).
exclusiveItems([]int{1, 2, 3}, []int{2, 3, 4}) // → [1, 4]
exclusiveItems([]int{1, 2}, []int{1, 2}) // → []
exclusiveItems([]int{5}, []int{6}) // → [5, 6]
This is the mirror image of the shared-elements problem: instead of values in both lists, you want values in exactly one. That's the symmetric difference — everything in one set or the other, but not both.
A value belongs in the answer if it's in the first list but missing from the second, or in the second but missing from the first. Both are membership questions, so reach for hash sets again: membership checks against a set are constant-time, while scanning a raw list is linear — and you'd be repeating that scan.
The two checks — "in first, not second" and "in second, not first" — can never both be true for the same value, so there's no risk of double-counting by running them as two independent passes.
Building both sets costs time proportional to the combined list sizes. Each pass costs time proportional to the set being walked, with constant-time lookups against the other. Total work stays linear in the combined input size, versus a brute-force comparison of every element against every other element, which costs time proportional to the product of the two sizes.
def exclusive_items(a, b):
sa = set(a)
sb = set(b)
res = [v for v in sa if v not in sb]
res += [v for v in sb if v not in sa]
res.sort()
return res
Sign in to join the discussion.
No comments yet. Be the first to ask a question.
Press Run to execute your code, or Submit to test and complete this problem.