2215. Find the Difference of Two Arrays

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

• answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
• answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

Understand the problem

The problem talks about unique values and “in this set but not that set,” which matches set semantics exactly: sets drop duplicates automatically and support a difference operator.

If you skip sets and check with if x in nums2 on a list, each check is O(n) and the whole solution becomes O(n²). Converting to sets first is worth it because membership checks become O(1).

Approach

1. Convert nums1 to set s1 and nums2 to set s2 (dedupes for free)
2. Compute s1 - s2 = values in s1 but not in s2
3. Compute s2 - s1 = values in s2 but not in s1
4. Return a two-element list, converting each set back to a list

Walkthrough — nums1 = [1, 2, 3, 3], nums2 = [1, 1, 2, 2]

1. Convert to sets: s1 = {1, 2, 3} (duplicate 3 collapses), s2 = {1, 2}
2. s1 − s2 = {1, 2, 3} − {1, 2} = {3}
3. s2 − s1 = {1, 2} − {1, 2, 3} = ∅ (empty)
4. Convert back to lists → [[3], []]

Edge case: identical arrays like nums1 = [1, 2], nums2 = [1, 2] yield empty sets both ways → [[], []], which is correct. Negative numbers work fine in a Python set.

Try it yourself first

Approach and walkthrough are above — write it yourself, then open the fold below when stuck or ready to compare.

def find_difference(nums1, nums2):
    s1, s2 = set(nums1), set(nums2)  # dedupe each array
    # s1 - s2 = in s1 but not in s2
    # s2 - s1 = in s2 but not in s1
    return [list(s1 - s2), list(s2 - s1)]

print(find_difference([1, 2, 3], [2, 4, 6]))        # [[1, 3], [4, 6]]
print(find_difference([1, 2, 3, 3], [1, 1, 2, 2]))  # [[3], []]

[[1, 3], [4, 6]]
[[3], []]