Fast Deduplication and Set Operations in Python

Question

Given two lists of customer IDs representing visitors to Campaign A and Campaign B, find all customers who visited both campaigns. Explain the time complexity versus nested loops.

Accepted Answer

Optimal Solution:
def find_common_customers(campaigns: dict) -> list[int]:
    list_a = campaigns['campaign_a']
    list_b = campaigns['campaign_b']
    # Set intersection is O(len(a) + len(b))
    return sorted(set(list_a).intersection(set(list_b)))

Technical Explanation:
Converting `list_a` to a set provides O(1) hash lookups. `set(list_a) & set(list_b)` runs in linear O(N + M) time compared to O(N * M) nested lists.