Solution: Redundant Connection
Let's solve the Redundant Connection problem using the Union Find pattern.
Statement
We’re given an undirected graph consisting of nodes. The graph is represented as an array called edges
, of length , where edges[i] = [a, b]
indicates that there is an edge between nodes a
and b
in the graph.
Return an edge that can be removed to make the graph a edges
.
Constraints:
edges.length
edges[i].length
2-
a
b
- There are no repeated edges.
- The given graph is connected.
- The graph contains only one cycle.
Solution
So far, you have probably brainstormed some approaches and have an idea of how to solve this problem. Let’s explore some of these approaches and figure out which one to follow based on considerations such as time complexity and any implementation constraints.
Naive approach
We can solve this problem using techniques like DFS or BFS, however, the naive approach using DFS or BFS has a time complexity of in the worst-case scenario. This is because the DFS or BFS algorithm will explore the graph by visiting each node and its edges, which may result in visiting many nodes multiple times, and not necessarily finding the redundant connection in an efficient manner.
Optimized approach using union find
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