Path Existence in an Undirected Graph

You are given an undirected graph with n vertices numbered from 0 to n - 1. The array edges contains one unordered pair [u, v] for each edge between vertices u and v.
Return true if a path connects source to destination. Otherwise, return false.
A vertex always has a path to itself, including an isolated vertex.

Function
validPath(n: int, edges: int[][], source: int, destination: int) → boolean

Examples
Example 1
n = 3
edges = [[0,1],[1,2],[2,0]]
source = 0
destination = 2
return = true
The edge between 0 and 2 directly connects the two vertices.
Example 2
n = 6
edges = [[0,1],[0,2],[3,5],[5,4],[4,3]]
source = 0
destination = 5
return = false
Vertices 0 and 5 belong to different connected components.
Example 3
n = 1
edges = []
source = 0
destination = 0
return = true
The source and destination are the same isolated vertex.

Constraints
1 <= n <= 200000.
0 <= edges.length <= 200000.
Every edge contains two valid vertex numbers.
0 <= source, destination < n.
The graph may contain cycles, disconnected components, and repeated edges.

Two clean ways to solve it. First, build an adjacency list, then run BFS or DFS from source with a visited array and return true the moment you reach destination. Second, use union-find: union every edge, then compare find(source) and find(destination). Both run in near-linear time. Pitfalls are predictable. Recursive DFS can blow the stack at 200000 nodes, so go iterative or use BFS. Forgetting to add edges in both directions breaks the undirected assumption. Skipping the visited check loops forever on cycles and repeated edges. Return true immediately if source equals destination, which covers Example 3 with n = 1 and no edges. Union-find with path compression avoids recursion trouble entirely. If you freeze during the live OA, StealthCoder can hand you the working template, but the pattern is simple enough to write from memory.

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