Is Minimize Maximum Component Cost easy or hard?

The problem is generally rated Medium because it requires recognizing the greedy pattern and implementing Union-Find correctly. Developers familiar with minimum spanning tree techniques usually identify the solution quickly. Beginners may struggle with component tracking and DSU optimizations.

Minimize Maximum Component Cost Python/Java solution

Most implementations follow the same pattern: sort edges, initialize a Union-Find structure, and iterate through edges performing union operations. Python typically uses a parent array with path compression, while Java implementations use class-based DSU structures. Both achieve O(E log E) time complexity.

How to solve Minimize Maximum Component Cost in O(E log E)?

Sort all edges by their cost and iterate through them in ascending order. Use a Union-Find structure to merge components whenever two vertices belong to different sets. Since edges are processed from smallest to largest, the maximum cost introduced into any component stays minimized.

What is the best approach for Minimize Maximum Component Cost?

The most effective approach is sorting edges by cost and merging components using Union-Find. This greedy strategy ensures smaller costs are processed first, preventing unnecessary increases in the maximum component cost. The method runs in O(E log E) time due to sorting and nearly constant-time union operations.

Is Minimize Maximum Component Cost asked at Google/Amazon/Meta?

Graph connectivity problems using Union-Find appear frequently in interviews at companies like Google, Amazon, and Meta. Variants often resemble minimum spanning tree or component-merging optimization tasks. Recognizing the Kruskal-style greedy pattern is commonly expected.

What data structure is used in Minimize Maximum Component Cost?

The core data structure is Union-Find (Disjoint Set Union). It tracks connected components efficiently and supports near-constant-time merging with path compression and union by rank. Sorting is also required to process edges in increasing cost order.

What is the time complexity of Minimize Maximum Component Cost?

The optimal solution runs in O(E log E) time because the edges must be sorted before processing. Union-Find operations such as union and find run in amortized O(α(V)) time with path compression and union by rank, making them effectively constant. Space complexity is O(V) for the disjoint-set structure.

Description Hints 2 Solution Video 19:50

3613. Minimize Maximum Component Cost

Medium43.5% acceptance4 companies

You are given an undirected connected graph with n nodes labeled from 0 to n - 1 and a 2D integer array edges where edges[i] = [u_i, v_i, w_i] denotes an undirected edge between node u_i and node v_i with weight w_i, and an integer k.

You are allowed to remove any number of edges from the graph such that the resulting graph has at most k connected components.

The cost of a component is defined as the maximum edge weight in that component. If a component has no edges, its cost is 0.

Return the minimum possible value of the maximum cost among all components after such removals.

Example 1:

Input: n = 5, edges = [[0,1,4],[1,2,3],[1,3,2],[3,4,6]], k = 2

Output: 4

Explanation:

Remove the edge between nodes 3 and 4 (weight 6).
The resulting components have costs of 0 and 4, so the overall maximum cost is 4.

Example 2:

Input: n = 4, edges = [[0,1,5],[1,2,5],[2,3,5]], k = 1

Output: 5

Explanation:

No edge can be removed, since allowing only one component (k = 1) requires the graph to stay fully connected.
That single component’s cost equals its largest edge weight, which is 5.

Constraints:

1 <= n <= 5 * 10⁴
0 <= edges.length <= 10⁵
edges[i].length == 3
0 <= u_i, v_i < n
1 <= w_i <= 10⁶
1 <= k <= n
The input graph is connected.

Asked at 4 companies

Salesforce Apple

Google Microsoft

Topics

Hints open one at a time

Hint 1

Sort the edges and do binary search on the candidate maximum weight

Hint 2

Use DFS or DSU to count the number of connected components when keeping only edges with weight <= mid

19:50

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1class Solution:
2    def minCost(self, n: int, edges: List[List[int]], k: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

5
[[0,1,4],[1,2,3],[1,3,2],[3,4,6]]
2

Input

4
[[0,1,5],[1,2,5],[2,3,5]]
1

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