How do you solve Minimize the Maximum Edge Weight of Graph on LeetCode?

Minimize the Maximum Edge Weight of Graph (LeetCode #3419) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(n) space complexity. Key insight: Binary search allows us to efficiently narrow down the maximum edge weight.

What is the brute force solution for Minimize the Maximum Edge Weight of Graph?

The brute force approach for Minimize the Maximum Edge Weight of Graph 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 find the maximum edge weight while ensuring node 0 is reachable from all other nodes. This method is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n log m).

What are the interview tips for Minimize the Maximum Edge Weight of Graph?

Clearly explain your thought process and why you choose each approach. Practice edge cases, especially with the threshold constraint. Be prepared to discuss the trade-offs of different algorithms.

What are common mistakes when solving Minimize the Maximum Edge Weight of Graph?

Failing to account for the threshold when adding edges to the graph. Not properly checking if all nodes can reach node 0 after filtering edges.

#3419

Minimize the Maximum Edge Weight of Graph

Medium

Binary SearchDepth-First SearchBreadth-First SearchGraph TheoryShortest PathBinary SearchGraph TraversalGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution uses binary search on the maximum edge weight combined with a graph traversal to check if we can reach node 0 under the weight constraint. This significantly reduces the number of combinations we need to check.

⚙️

Algorithm

3 steps

1Step 1: Sort the edges based on their weights.
2Step 2: Use binary search on the edge weights to find the minimum maximum weight that allows all nodes to reach node 0.
3Step 3: For each mid value in the binary search, create a subgraph with edges that have weights less than or equal to mid, and check if node 0 is reachable from all other nodes using BFS or DFS.

solution.py34 lines

1# Full working Python code
2from collections import defaultdict, deque
3
4def can_reach_zero(n, graph):
5    visited = [False] * n
6    queue = deque([0])
7    while queue:
8        node = queue.popleft()
9        visited[node] = True
10        for neighbor in graph[node]:
11            if not visited[neighbor]:
12                queue.append(neighbor)
13    return all(visited)
14
15
16def min_max_edge_weight(n, edges, threshold):
17    edges.sort(key=lambda x: x[2])
18    left, right = 0, edges[-1][2]
19    while left < right:
20        mid = (left + right) // 2
21        graph = defaultdict(list)
22        outgoing = [0] * n
23        for u, v, w in edges:
24            if w <= mid:
25                graph[u].append(v)
26                outgoing[u] += 1
27            if outgoing[u] > threshold:
28                break
29        else:
30            if can_reach_zero(n, graph):
31                right = mid
32            else:
33                left = mid + 1
34    return left if left <= edges[-1][2] else -1

ℹ

Complexity note: The time complexity is O(n log m) where n is the number of nodes and m is the number of edges. The binary search runs log m times, and each reachability check runs in O(n) time.

1Binary search allows us to efficiently narrow down the maximum edge weight.
2Graph traversal techniques (DFS/BFS) are essential for checking reachability.

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