How do you solve Minimum Cost Walk in Weighted Graph on LeetCode?

Minimum Cost Walk in Weighted Graph (LeetCode #3108) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding the bitwise AND operation is crucial for determining the cost of walks.

What is the time complexity of Minimum Cost Walk in Weighted Graph?

The optimal solution for Minimum Cost Walk in Weighted Graph runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the union-find operations, which are nearly constant time due to path compression and union by rank.

What is the brute force solution for Minimum Cost Walk in Weighted Graph?

The brute force approach for Minimum Cost Walk in Weighted Graph is Brute Force, with O(n²) time complexity and O(n) space complexity. We can explore all possible walks from the starting vertex to the target vertex. For each walk, we calculate the bitwise AND of the weights of the edges traversed. This approach guarantees that we will find the minimum cost, but it is inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Cost Walk in Weighted Graph?

Minimum Cost Walk in Weighted Graph uses the following concepts: Array, Bit Manipulation, Union-Find, Graph Theory. The recommended patterns to study are: Disjoint Set Union (Union-Find), Graph Traversal.

What are the interview tips for Minimum Cost Walk in Weighted Graph?

Always clarify the problem constraints and edge cases with the interviewer. Explain your thought process clearly while coding to demonstrate your understanding. Practice implementing the union-find structure, as it is a common pattern in graph-related problems.

What are common mistakes when solving Minimum Cost Walk in Weighted Graph?

Failing to consider that a walk can revisit vertices and edges. Not properly implementing the union-find structure, leading to incorrect component identification.

#3108

Minimum Cost Walk in Weighted Graph

Hard

ArrayBit ManipulationUnion-FindGraph TheoryDisjoint Set Union (Union-Find)Graph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

Using the Disjoint Set Union (DSU) allows us to efficiently group connected components. We can then determine if two vertices are connected and calculate the minimum cost of the walk based on the weights of the edges in the component.

⚙️

Algorithm

3 steps

1Step 1: Create a Disjoint Set Union (DSU) to group connected components based on the edges.
2Step 2: For each edge, union the two vertices and keep track of the minimum weight in each component.
3Step 3: For each query, check if the two vertices belong to the same component using DSU. If they do, return the minimum weight; otherwise, return -1.

solution.py39 lines

1# Full working Python code
2class DSU:
3    def __init__(self, n):
4        self.parent = list(range(n))
5        self.rank = [0] * n
6        self.min_weight = [float('inf')] * n
7
8    def find(self, u):
9        if self.parent[u] != u:
10            self.parent[u] = self.find(self.parent[u])
11        return self.parent[u]
12
13    def union(self, u, v, weight):
14        root_u = self.find(u)
15        root_v = self.find(v)
16        if root_u != root_v:
17            if self.rank[root_u] > self.rank[root_v]:
18                self.parent[root_v] = root_u
19                self.min_weight[root_u] = min(self.min_weight[root_u], self.min_weight[root_v], weight)
20            elif self.rank[root_u] < self.rank[root_v]:
21                self.parent[root_u] = root_v
22                self.min_weight[root_v] = min(self.min_weight[root_u], self.min_weight[root_v], weight)
23            else:
24                self.parent[root_v] = root_u
25                self.min_weight[root_u] = min(self.min_weight[root_u], self.min_weight[root_v], weight)
26                self.rank[root_u] += 1
27
28
29def min_cost_walk(n, edges, queries):
30    dsu = DSU(n)
31    for u, v, w in edges:
32        dsu.union(u, v, w)
33    results = []
34    for s, t in queries:
35        if dsu.find(s) == dsu.find(t):
36            results.append(dsu.min_weight[dsu.find(s)])
37        else:
38            results.append(-1)
39    return results

ℹ

Complexity note: The time complexity is O(n log n) due to the union-find operations, which are nearly constant time due to path compression and union by rank.

1Understanding the bitwise AND operation is crucial for determining the cost of walks.
2Using a Disjoint Set Union (DSU) can significantly optimize the process of finding connected components.

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