The example they gave made it click pretty fast: edges like [0,1],[0,2],[3,4] give you groups of size 3 and 2, so 3x2=6 pairs.
Model the employees as a graph where edges represent connections, then use Union-Find (Disjoint Set Union) to efficiently group employees into connected components. After identifying the sizes of each component, compute the total number of cross-group pairs by summing the products of sizes of all pairs of distinct components, or equivalently using the formula (total^2 - sum(size_i^2))/2.
Pro tip: Mention that Union-Find with path compression and union by rank gives near O(E) time, which is optimal for this problem. Also, clarify that the pair count can be computed in O(N) after components are found, avoiding O(N^2) enumeration.
Confirm that edges are undirected and that employees are identified by integers or strings. Ask about isolated employees (no edges) and whether they should form their own groups.
Select Union-Find (Disjoint Set Union) for efficient connected component detection. Explain that it supports near-constant time union and find operations with path compression and union by rank.
Iterate through all edges, union the two endpoints. After processing, traverse all employees to find their root and count the size of each component.
Use the formula: total pairs = (total_employees^2 - sum(size_i^2)) / 2. Alternatively, accumulate pairs by multiplying the current component size with the sum of sizes of previously processed components.
State that time complexity is O(E α(N)) for Union-Find and O(N) for counting, overall near O(E+N). Test with small cases, disconnected graphs, and a single component.
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