How do you solve Redundant Connection on LeetCode?

Redundant Connection (LeetCode #684) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n α(n)) time complexity and O(n) space complexity. Key insight: Using Union-Find is efficient for cycle detection in graphs.

What is the brute force solution for Redundant Connection?

The brute force approach for Redundant Connection 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 edges can still connect all nodes without forming a cycle. The optimal Optimal Solution approach improves this to O(n α(n)).

What are the interview tips for Redundant Connection?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches. Practice implementing Union-Find as it's a common pattern in graph problems.

What are common mistakes when solving Redundant Connection?

Not correctly implementing the Union-Find structure. Failing to handle the edge cases where multiple redundant edges exist.

#684

Redundant Connection

Medium

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

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n α(n))Space O(n)

The optimal solution uses the Union-Find (Disjoint Set Union) data structure to efficiently detect cycles. By iterating through the edges and attempting to union the nodes, we can quickly determine if adding an edge would create a cycle.

⚙️

Algorithm

4 steps

1Step 1: Initialize a Union-Find structure to manage connected components.
2Step 2: Iterate through each edge in the graph.
3Step 3: For each edge, check if the two nodes are already connected using the Union-Find structure.
4Step 4: If they are connected, this edge is the redundant one. If not, union the two nodes.

solution.py29 lines

1class UnionFind:
2    def __init__(self, size):
3        self.parent = list(range(size))
4        self.rank = [1] * size
5
6    def find(self, u):
7        if self.parent[u] != u:
8            self.parent[u] = self.find(self.parent[u])
9        return self.parent[u]
10
11    def union(self, u, v):
12        rootU = self.find(u)
13        rootV = self.find(v)
14        if rootU == rootV:
15            return False
16        if self.rank[rootU] > self.rank[rootV]:
17            self.parent[rootV] = rootU
18        elif self.rank[rootU] < self.rank[rootV]:
19            self.parent[rootU] = rootV
20        else:
21            self.parent[rootV] = rootU
22            self.rank[rootU] += 1
23        return True
24
25def findRedundantConnection(edges):
26    uf = UnionFind(len(edges) + 1)
27    for u, v in edges:
28        if not uf.union(u, v):
29            return [u, v]

ℹ

Complexity note: The time complexity is O(n α(n)), where α is the Inverse Ackermann function, which grows very slowly, making this approach nearly linear. The space complexity is O(n) due to the storage of the parent and rank arrays.

1Using Union-Find is efficient for cycle detection in graphs.
2Understanding the properties of trees helps in identifying redundant edges.

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