Problem Description

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.

Intuition

The goal is to break the graph into at most k connected components while minimizing the largest maximum edge weight in any component. Removing edges generally increases the number of components, and when a component is split, removing a large-weight edge will directly affect the cost in the resulting components.

If we sort all edges by their weights from smallest to largest, we can think of the problem as keeping the graph as connected as possible with the smallest possible maximum edge weight within each component. The fewer edges we keep, the more components we have.

Using a union-find data structure, we can build the graph by adding edges from smallest to largest weight. Each time an edge connects two components, the total number of components decreases. Once we have exactly k components, the last edge's weight we added determines the maximum edge weight in any component. This is because all remaining edges are larger and can be removed. If k equals n, the graph can be fully disconnected into single nodes, so the maximum cost is 0.

Solution Approach

The solution uses sorting and the union-find (disjoint set) data structure. Here’s how the approach works:

1. Special Case (k == n): If k equals n, every node can become its own component, with no edges left. In this case, the cost of every component is 0, so we can directly return 0.

2. Sort Edges: Sort edges based on their weights in ascending order. This ensures that lighter edges (with smaller weights) are considered first. Using smaller weighted edges to connect the components helps keep the maximum cost low.

3. Union-Find Initialization: Prepare a union-find data structure by assigning each node as its own parent:

   p = list(range(n))

   The find function applies path compression to efficiently determine the parent of each node.

4. Build Components by Connecting Edges: Start connecting nodes using the sorted edges. For each edge [u, v, w], connect u and v if they are currently in different components. This merges two components into one.

   Every successful merge decreases the component count (cnt -= 1). Once the number of components is reduced to k, the current edge's weight (w) is the answer. This is because, from this point on, any further edges added (which are heavier) would not be needed, and removing them or not adding them won't increase the number of components beyond k.

5. Return the Result: Once cnt == k, return the weight w of the current edge. This represents the minimum possible value for the maximum cost among all components.

Key patterns used:

• Greedy: Always connect components with the smallest possible edge to avoid unnecessarily high costs.
• Union-Find: Efficiently manage and merge connected components.

Example in code context:

edges.sort(key=lambda x: x[2])
p = list(range(n))
cnt = n
for u, v, w in edges:
    pu, pv = find(u), find(v)
    if pu != pv:
        p[pu] = pv
        cnt -= 1
        if cnt <= k:
            return w

This ensures the answer is found as soon as the number of components does not exceed k. If cnt never becomes k or less (which shouldn't happen given the problem constraints), the fallback is returning 0.

Example Walkthrough

Suppose we have a graph with n = 5 nodes and the following edges:

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

Objective: Remove any number of edges so that there are at most k = 2 connected components, and the largest maximum edge weight in any component is minimized.

Step 1. Sort Edges by Weight: Sorted edges: [ [0, 1, 2], [1, 2, 3], [0, 2, 4], [1, 3, 5], [2, 4, 6] ]

Step 2. Initialize Union-Find: Each node is its own component: p = [0, 1, 2, 3, 4] Number of components: cnt = 5

Step 3. Iterate Through Edges:

• Add edge [0, 1, 2]:
  • Connect 0 and 1.
  • Components: 4
• Add edge [1, 2, 3]:
  • Connect (0-1) and 2.
  • Components: 3
• Add edge [0, 2, 4]:
  • 0 and 2 are already connected. Skip.
• Add edge [1, 3, 5]:
  • Connect (0-1-2) and 3.
  • Components: 2

At this point (cnt = k = 2), the last edge added had a weight of 5.

Step 4. Return the Result: The answer is 5, which is the minimum possible value of the maximum cost among all components after breaking the graph into at most 2 connected components.

Alternatives:

• Removing heavier edges first could result in a greater maximum cost.
• If k = 3, the process would stop after adding edge [1, 2, 3], with the answer being 3.

Summary Table:

Final Components:

• One component: {0, 1, 2, 3} (max edge 5)
• One component: {4} (no edge, cost 0)

Result: Minimum possible value of the maximum cost among all components is 5.

Solution Implementation

Time and Space Complexity

• Time Complexity: The dominant operations are sorting the edges list and performing union-find operations.

  • Sorting the edges takes O(m \log m), where m is the number of edges.
  • Each union-find operation with path compression is nearly O(1) amortized, and at most O(m) such operations occur. So, the overall time complexity is O(m \log m). If the graph is connected and dense, m can be up to O(n^2), but usually m is the input size.

• Space Complexity: The space complexity comes from the parent array p of size n, so it is O(n).

Thus, compared to the reference answer which states O(n \log n) time, the compiled analysis is:

• Time complexity: O(m \log m) (where m is number of edges)
• Space complexity: O(n) (where n is number of nodes)