Minimum Score of a Path Between Cities

You are given n cities numbered from 1 through n and an array roads. Each roads[i] = [a, b, distance] describes a bidirectional road between cities a and b.
The score of a path is the minimum road distance used anywhere along that path. Cities and roads may be visited more than once.
Return the minimum possible score of a path from city 1 to city n. At least one such path exists.

Function
minScore(n: int, roads: int[][]) → int

Examples
Example 1
n = 4
roads = [[1,2,9],[2,3,6],[2,4,5],[1,4,7]]
return = 5
The path 1 -> 2 -> 4 has score min(9, 5) = 5, which is optimal.
Example 2
n = 4
roads = [[1,2,2],[1,3,4],[3,4,7]]
return = 2
Because revisits are allowed, a walk can use the road of length 2 before continuing to city 4.

Here's the trick. Revisits are allowed, so any road reachable from city 1 can be walked to and walked back from. Since a path from 1 to n is guaranteed, every edge in the component containing city 1 (and therefore n) can be part of some valid walk. The answer is the minimum edge weight in that component. Run a BFS or DFS from city 1 over an adjacency list, and track the smallest distance across every edge you touch. Union-find works too: union all edges, then take the minimum weight among edges whose endpoint shares city 1's root. The common pitfall is running Dijkstra or hunting for the best simple path, which wastes time and gets it wrong. Another is forgetting roads are bidirectional, so you must add both directions. Complexity is O(n + m). If you freeze on the reduction during the live OA, StealthCoder can surface the component-minimum approach quickly.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.