Minimum Cost for Roads and Hospitals

Cities are labeled from 0 to cityCount - 1. A hospital can be built in any city for hospitalCost, and any listed undirected road can be built for roadCost.
Choose hospitals and roads at minimum total cost so every city can reach a hospital.

Function
minimumInfrastructureCost(cityCount: int, possibleRoads: int[][], hospitalCost: int, roadCost: int) → int

Examples
Example 1
cityCount = 3
possibleRoads = [[0,1],[1,2]]
hospitalCost = 3
roadCost = 2
return = 7
Case 1 exercises the documented deterministic contract.
Example 2
cityCount = 3
possibleRoads = [[0,1]]
hospitalCost = 2
roadCost = 5
return = 6
Case 2 exercises the documented deterministic contract.
Example 3
cityCount = 5
possibleRoads = [[0,1],[2,3]]
hospitalCost = 6
roadCost = 1
return = 20
Case 3 exercises the documented deterministic contract.

Constraints
1 <= cityCount <= 200000.
0 <= possibleRoads.length <= 200000.
Costs are positive and the answer fits a signed 32-bit integer.

Here's what it reduces to: every connected component needs exactly one hospital. Inside a component of size k, you can connect all cities with k-1 roads (a spanning tree), or skip roads and build a hospital in each city for k * hospitalCost. So per component, cost is hospitalCost + (k-1) * min(roadCost, hospitalCost). Sum across components. Check example 3: components sizes 2, 2, 1. Hospital 6, road 1. Two pairs cost 6+1 = 7 each, the singleton costs 6. Total 20. Correct. Use union-find or iterative BFS, since cityCount reaches 200000 and recursive DFS can overflow the stack. The common pitfall is always building roads even when roadCost exceeds hospitalCost, as in example 2. Another is forgetting isolated cities, which are their own components. If you freeze in the live OA, StealthCoder can hand you the union-find skeleton, but the formula is the real trick.

StealthCoder is the hedge for the one pattern you didn't drill. It runs invisibly during the screen share.