How do you solve Redundant Connection II on LeetCode?

Redundant Connection II (LeetCode #685) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using Union-Find helps efficiently detect cycles in directed graphs.

What is the brute force solution for Redundant Connection II?

The brute force approach for Redundant Connection II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each edge to see if removing it results in a valid tree. This means we will try removing each edge one by one and check if the remaining graph is a valid rooted tree. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Redundant Connection II?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as when there are multiple redundant connections. Practice explaining your thought process while coding to simulate the interview environment.

What are common mistakes when solving Redundant Connection II?

Failing to properly implement cycle detection in the graph. Not considering the order of edges when multiple solutions exist.

#685

Redundant Connection II

Hard

Depth-First SearchBreadth-First SearchUnion-FindGraph TheoryUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses a Union-Find (Disjoint Set Union) data structure to efficiently manage and check cycles in the graph. We can keep track of the parent of each node and detect if adding an edge creates a cycle.

⚙️

Algorithm

3 steps

1Step 1: Initialize a parent array to keep track of the parent of each node.
2Step 2: Iterate through each edge and try to union the nodes. If both nodes already have a parent, mark this edge as a candidate for removal.
3Step 3: If a cycle is detected during the union operation, return the last edge that caused the cycle.

solution.py13 lines

1def findRedundantDirectedConnection(edges):
2    parent = [0] * (len(edges) + 1)
3    def find(x):
4        if parent[x] == 0:
5            return x
6        parent[x] = find(parent[x])
7        return parent[x]
8    for u, v in edges:
9        pu, pv = find(u), find(v)
10        if pu == pv:
11            return [u, v]
12        parent[pu] = pv
13    return []

ℹ

Complexity note: The time complexity is O(n) because we perform a union-find operation for each edge, which is efficient due to path compression. The space complexity is O(n) for the parent array.

1Using Union-Find helps efficiently detect cycles in directed graphs.
2Understanding the properties of trees is crucial for solving graph-related problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.