What is the time complexity of Remove Max Number of Edges to Keep Graph Fully Traversable?

The optimal solution for Remove Max Number of Edges to Keep Graph Fully Traversable runs in O(n log n) time and uses O(n) space. The sorting step dominates the time complexity, while the Union-Find operations are nearly constant time due to path compression.

What is the brute force solution for Remove Max Number of Edges to Keep Graph Fully Traversable?

The brute force approach for Remove Max Number of Edges to Keep Graph Fully Traversable is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of edges to see if the graph remains fully traversable after removing them. This is straightforward but inefficient, as it doesn't scale well with larger graphs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Remove Max Number of Edges to Keep Graph Fully Traversable?

Remove Max Number of Edges to Keep Graph Fully Traversable uses the following concepts: Union-Find, Graph Theory. The recommended patterns to study are: Union-Find, Graph Traversal.

What are the interview tips for Remove Max Number of Edges to Keep Graph Fully Traversable?

Explain your thought process clearly while coding. Discuss edge cases and how your solution handles them. Practice Union-Find problems to become comfortable with the data structure.

What are common mistakes when solving Remove Max Number of Edges to Keep Graph Fully Traversable?

Not considering the order of edge types when processing. Failing to check connectivity after processing edges.

#1579

Remove Max Number of Edges to Keep Graph Fully Traversable

Hard

Union-FindGraph TheoryUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses a Union-Find (Disjoint Set Union) data structure to efficiently manage connectivity between nodes. By prioritizing edges that both Alice and Bob can traverse, we can ensure maximum edge removal while maintaining full connectivity.

⚙️

Algorithm

4 steps

1Step 1: Sort edges by type, prioritizing type 3 edges first, then type 1, and finally type 2.
2Step 2: Use Union-Find to connect nodes based on the edges.
3Step 3: Count the number of edges used and check if both Alice and Bob can traverse all nodes.
4Step 4: Calculate the maximum removable edges based on the total edges and the edges used.

solution.py51 lines

1# Full working Python code
2class UnionFind:
3    def __init__(self, size):
4        self.parent = list(range(size))
5        self.rank = [1] * size
6
7    def find(self, u):
8        if self.parent[u] != u:
9            self.parent[u] = self.find(self.parent[u])
10        return self.parent[u]
11
12    def union(self, u, v):
13        rootU = self.find(u)
14        rootV = self.find(v)
15        if rootU != rootV:
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
24
25def max_edges_to_remove(n, edges):
26    ufAlice = UnionFind(n + 1)
27    ufBob = UnionFind(n + 1)
28    totalEdges = len(edges)
29    usedEdges = 0
30
31    # Sort edges by type
32    edges.sort(key=lambda x: -x[0])
33
34    for edge in edges:
35        if edge[0] == 3:
36            ufAlice.union(edge[1], edge[2])
37            ufBob.union(edge[1], edge[2])
38            usedEdges += 1
39        elif edge[0] == 1:
40            ufAlice.union(edge[1], edge[2])
41            usedEdges += 1
42        elif edge[0] == 2:
43            ufBob.union(edge[1], edge[2])
44            usedEdges += 1
45
46    # Check if both can traverse
47    if ufAlice.find(1) != ufAlice.find(n) or ufBob.find(1) != ufBob.find(n):
48        return -1
49
50    return totalEdges - usedEdges
51

ℹ

Complexity note: The sorting step dominates the time complexity, while the Union-Find operations are nearly constant time due to path compression.

1Prioritize edges that can be traversed by both Alice and Bob to maximize removable edges.
2Use Union-Find to efficiently manage connectivity between nodes.

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