Problem Description

You have an undirected connected graph with n nodes labeled from 0 to n - 1. The graph is defined by a 2D array edges where each edges[i] = [ui, vi, wi] represents an undirected edge between nodes ui and vi with weight wi. You're also given an integer k.

Your task is to remove some edges from the graph so that the resulting graph has at most k connected components.

After removing edges, each connected component will have a cost, which is defined as the maximum edge weight within that component. If a component has no edges (it's just an isolated node), its cost is 0.

Your goal is to minimize the maximum cost among all components after removing edges.

For example, if after removing edges you have 3 components with costs [5, 8, 2], the maximum cost would be 8. You want to find a way to remove edges such that this maximum cost is as small as possible.

The problem asks you to return the minimum possible value of the maximum cost among all components after optimal edge removals.

Intuition

The key insight is to think about this problem in reverse. Instead of removing edges from a connected graph, we can start with all nodes isolated and gradually add edges to form components.

If we're allowed to have k components, we need to reduce the number of components from n (all isolated) down to k. This means we need to add exactly n - k edges to merge components together.

Here's the crucial observation: to minimize the maximum edge weight in any component, we should prioritize adding edges with smaller weights first. Why? Because once we add an edge with weight w, at least one component will have a maximum edge weight of at least w.

So the strategy becomes:

1. Sort all edges by weight in ascending order
2. Start with n isolated nodes (components)
3. Add edges one by one from smallest to largest weight
4. Each edge that connects two different components reduces the total component count by 1
5. Stop when we reach exactly k components

The weight of the last edge we add will be our answer. This is because:

• All edges with smaller weights have already been added (and are distributed among the components)
• All edges with larger weights can be removed (we don't need them to maintain k components)
• The last edge we add represents the minimum possible maximum edge weight we must tolerate

Special case: If k = n, we can remove all edges, leaving all nodes isolated. In this case, the maximum cost is 0 since isolated nodes have no edges.

Learn more about Union Find, Graph, Binary Search and Sorting patterns.

Solution Approach

The solution uses Sorting combined with Union-Find (Disjoint Set Union) data structure to efficiently solve this problem.

Step 1: Handle Special Case

if k == n:
    return 0

If k equals n, we can have all nodes as separate components (remove all edges), so the maximum cost is 0.

Step 2: Sort Edges by Weight

edges.sort(key=lambda x: x[2])

We sort all edges in ascending order by their weights. This allows us to process edges from smallest to largest weight.

Step 3: Initialize Union-Find Structure

cnt = n
p = list(range(n))

• cnt tracks the current number of connected components (initially n since all nodes are isolated)
• p is the parent array for Union-Find, where p[i] initially points to itself (each node is its own parent)

Step 4: Union-Find Find Operation

def find(x: int) -> int:
    if p[x] != x:
        p[x] = find(p[x])  # Path compression
    return p[x]

This function finds the root parent of node x with path compression optimization, which flattens the tree structure for faster future lookups.

Step 5: Process Edges and Union Components

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

For each edge (u, v, w):

• Find the root parents of nodes u and v
• If they have different parents (pu != pv), they're in different components:
  • Union them by setting p[pu] = pv
  • Decrease the component count by 1
  • If we've reached k or fewer components, return the current edge weight w

The algorithm works because:

• We process edges from smallest to largest weight
• We only add an edge if it connects two different components
• The moment we reach k components, the current edge weight is the minimum possible maximum weight
• All remaining edges with larger weights can be safely removed

Time Complexity: O(E log E + E × α(n)) where E is the number of edges and α is the inverse Ackermann function (practically constant) Space Complexity: O(n) for the Union-Find parent array

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Given:

• n = 4 nodes (labeled 0, 1, 2, 3)
• edges = [[0,1,3], [1,2,5], [2,3,2], [0,3,4]]
• k = 2 (we want at most 2 connected components)

Step 1: Check special case

• Since k = 2 and n = 4, we proceed (k ≠ n)

Step 2: Sort edges by weight After sorting: edges = [[2,3,2], [0,1,3], [0,3,4], [1,2,5]]

Step 3: Initialize Union-Find

• Start with 4 isolated nodes: cnt = 4
• Parent array: p = [0, 1, 2, 3] (each node is its own parent)
• Visual representation:

0   1   2   3  (4 components)

Step 4: Process edges from smallest to largest

Edge 1: [2,3,2] with weight 2

• Find parents: find(2) = 2, find(3) = 3
• Different parents, so union them: p[2] = 3
• Component count: cnt = 3
• Visual:

0   1   2-3  (3 components)

• Check: cnt = 3 > k = 2, continue

Edge 2: [0,1,3] with weight 3

• Find parents: find(0) = 0, find(1) = 1
• Different parents, so union them: p[0] = 1
• Component count: cnt = 2
• Visual:

0-1   2-3  (2 components)

• Check: cnt = 2 = k, return weight 3

Result: The minimum possible maximum cost is 3.

Why this works:

• We formed exactly 2 components: {0,1} and {2,3}
• Component {0,1} has edge weight 3
• Component {2,3} has edge weight 2
• The maximum cost among components is 3
• We didn't need to add edges [0,3,4] or [1,2,5], which would have created larger maximum weights

If we had added edge [0,3,4] instead, it would merge all nodes into one component, violating our constraint of having at most 2 components. By stopping at weight 3, we achieve the minimum possible maximum cost while maintaining exactly k=2 components.

Solution Implementation

Time and Space Complexity

The time complexity of this code is O(m × log m + m × α(n)), where m is the number of edges and α(n) is the inverse Ackermann function. However, since α(n) grows extremely slowly and is effectively constant for all practical purposes, and considering that in the worst case m can be O(n²) for a complete graph, the time complexity can be simplified to O(n² × log n) in the worst case. But if we consider the number of edges as the primary input size, it would be O(m × log m).

Breaking down the operations:

• Sorting the edges: O(m × log m) where m = len(edges)
• Union-Find operations: Each find operation takes O(α(n)) amortized time with path compression, and we perform at most m union operations
• The loop iterates through all edges once: O(m)

The reference answer states O(n × log n), which appears to be considering a sparse graph where the number of edges m = O(n), making the sorting operation O(n × log n).

The space complexity is O(n) where n is the number of nodes:

• The parent array p stores n elements: O(n)
• The recursion stack for the find function in the worst case (before path compression) could be O(n), but with path compression, it becomes O(log n) on average
• Other variables use constant space: O(1)

Therefore, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Problem Goal

Pitfall: Many people initially think we need to create exactly k components, but the problem asks for "at most k" components. This confusion can lead to incorrect termination conditions.

Solution: The algorithm correctly handles this by returning as soon as component_count <= k, not waiting for exactly k components.

2. Forgetting Path Compression in Union-Find

Pitfall: Implementing the find operation without path compression:

# Incorrect - No path compression
def find(x):
    while parent[x] != x:
        x = parent[x]
    return x

This leads to degraded performance from O(α(n)) to O(n) per operation in worst case.

Solution: Always implement path compression:

def find(x):
    if parent[x] != x:
        parent[x] = find(parent[x])  # Recursively compress path
    return parent[x]

3. Incorrect Edge Case Handling for k = n

Pitfall: Forgetting to handle the special case where k == n. Without this check, the algorithm might process all edges unnecessarily or return an incorrect result.

Solution: Add the special case check at the beginning:

if k == n:
    return 0  # All nodes can be isolated, no edges needed

4. Using Wrong Union Strategy

Pitfall: Some implementations might try to always make the smaller indexed node the parent:

# Potentially problematic
if parent_source < parent_destination:
    parent[parent_destination] = parent_source
else:
    parent[parent_source] = parent_destination

While this works, it's unnecessary complexity. The direction of union doesn't affect correctness here.

Solution: Keep it simple - consistently union in one direction:

parent[parent_source] = parent_destination

5. Misinterpreting the Return Value

Pitfall: Returning the wrong value when the target is reached. Some might return the index, the next weight, or accumulate weights:

# Wrong - returning accumulated weight
total_weight += weight
if component_count <= k:
    return total_weight

Solution: Return the current edge weight when reaching k components, as this represents the maximum edge weight needed to form at most k components:

if component_count <= k:
    return weight  # This is the minimum possible maximum weight