Minimum-Cost Meeting City

An undirected, unweighted graph represents n cities numbered from 0 to n - 1. Travelers start at cities x and y. They may travel separately to a meeting city m, then ride together from m to destination.
The cost of choosing m is dist(x, m) + dist(y, m) + dist(m, destination), so the shared final route is counted once.
Return the meeting city with minimum cost. If several cities have the same minimum cost, return the smallest city index. Return -1 if no city is reachable from both travelers and can reach destination.

Function
minimumCostMeetingCity(n: int, edges: int[][], x: int, y: int, destination: int) → int

Examples
Example 1
n = 7
edges = [[0,2],[1,2],[2,3],[3,6],[0,4],[1,4],[4,5],[5,6]]
x = 0
y = 1
destination = 6
return = 2
Meeting at city 2 costs 1 + 1 + 2 = 4. Meeting at city 4 also costs 4, so the smaller city index 2 wins the tie.
Example 2
n = 4
edges = [[0,1],[2,3]]
x = 0
y = 2
destination = 3
return = -1
The travelers start in different connected components, so no valid meeting city exists.

Constraints
1 <= n <= 200000
0 <= edges.length <= 200000
Every row of edges is [u, v] with 0 <= u, v < n and u != v.
0 <= x, y, destination < n

Build an adjacency list from edges. Run BFS three times, from x, from y, and from destination. Because the graph is undirected and unweighted, BFS gives exact shortest distances in O(n + e). Then scan cities 0 to n-1 in order. Skip any city where one of the three distances is unreachable. Track the minimum sum, and only replace the best on a strictly smaller cost, so ties keep the smaller index automatically. The common pitfall is running BFS from every candidate city, which is O(n * (n + e)) and times out at 200000. Another one is forgetting that dist(m, destination) equals dist(destination, m) here because edges are undirected. Use -1 as the unreachable marker and don't add it into sums. If you freeze live, StealthCoder is the hedge that reads the problem and hands you this structure. Example 2 checks the disconnected case.

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