LeetCode 3613 - Minimize Maximum Component Cost

Let's dive into a full technical guide for LeetCode 3613 following your requested structure. The problem presents an undirected, connected graph with n nodes and weighted edges. Each edge connects two nodes and has a weight representing its "cost.

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LeetCode Problem 3613

Difficulty: 🟡 Medium
Topics: Binary Search, Union-Find, Graph Theory, Sorting

Solution

Let's dive into a full technical guide for LeetCode 3613 following your requested structure.

Problem Understanding

The problem presents an undirected, connected graph with n nodes and weighted edges. Each edge connects two nodes and has a weight representing its "cost." You are allowed to remove edges to create up to k connected components. The cost of a component is the maximum edge weight in that component, and the overall cost is the maximum of these component costs. The goal is to minimize this overall cost.

In simpler terms, imagine a graph as a network of islands connected by bridges of varying strength. You can destroy some bridges to split the network into smaller groups of islands (at most k groups). The strength of the strongest bridge in each group determines that group's cost, and the largest group cost is the "maximum cost." You want to strategically remove bridges so that the strongest bridge in any group is as weak as possible.

The input consists of n (number of nodes), edges (list of [u, v, w] triples representing undirected edges with weight w), and k (maximum allowed components). The output is a single integer, the minimum possible maximum component cost.

Important constraints: n can be up to 50,000 and edges up to 100,000, so brute-force enumeration of all edge removals is infeasible. The graph is guaranteed to be connected, so initially there is exactly one component.

Edge cases include graphs with no removable edges (k = 1), graphs where k = n (potentially removing almost all edges), and graphs where multiple edges share the same weight.

Approaches

Brute Force

The brute-force method would enumerate every possible combination of edge removals, compute the connected components after each removal, and calculate the maximum cost per configuration. Then we pick the configuration that yields the smallest maximum cost. While correct, this approach is infeasible due to combinatorial explosion: with m edges, there are 2^m possible removal sets, making it far too slow for m = 10^5.

Key Insight for Optimal Solution

Notice that the maximum component cost is entirely determined by the edges we keep, and the minimum maximum component cost will always correspond to edges in a maximum spanning forest. If we sort edges by weight in descending order and try to greedily include edges while forming components, we can use binary search over possible costs and a Union-Find structure to check if the graph can be partitioned into at most k components under a given cost threshold.

The observation is: for a candidate maximum cost C, we can remove all edges with weight greater than C and see how many components remain. If the count is ≤ k, C is feasible. Otherwise, C is too low. This enables binary search.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^m * (n + m)) O(n + m) Enumerates all edge removals and computes components each time
Optimal O(m log W + α(n)) O(n) Binary search on cost + Union-Find; α(n) is inverse Ackermann function for union operations

Algorithm Walkthrough

  1. Initialize Search Range: Set left = 0 and right = max_edge_weight. These define the possible maximum costs.
  2. Binary Search Loop: While left < right, pick a mid-point mid = (left + right) // 2. This represents a candidate maximum component cost.
  3. Simulate Edge Removals: Create a Union-Find data structure for n nodes. Iterate over all edges. Only union nodes if edge_weight <= mid. This effectively removes all edges with weight greater than mid.
  4. Count Components: After unioning valid edges, count the number of connected components.
  5. Adjust Search Range: If the number of components ≤ k, it is feasible to have mid as the maximum component cost, so we try to lower it by setting right = mid. Otherwise, set left = mid + 1 to increase the candidate cost.
  6. Return Result: After the loop, left will converge to the minimum feasible maximum component cost.

Why it works: At each iteration, we check whether a candidate cost can produce ≤ k components. Binary search ensures we find the smallest such feasible cost, and Union-Find guarantees correct component counting. By only including edges ≤ candidate cost, we directly model the effect of removing heavy edges.

Python Solution

from typing import List

class UnionFind:
    def __init__(self, n: int):
        self.parent = list(range(n))
        self.rank = [0] * n
        self.count = n

    def find(self, x: int) -> int:
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x: int, y: int) -> bool:
        xr, yr = self.find(x), self.find(y)
        if xr == yr:
            return False
        if self.rank[xr] < self.rank[yr]:
            self.parent[xr] = yr
        elif self.rank[xr] > self.rank[yr]:
            self.parent[yr] = xr
        else:
            self.parent[yr] = xr
            self.rank[xr] += 1
        self.count -= 1
        return True

class Solution:
    def minCost(self, n: int, edges: List[List[int]], k: int) -> int:
        left, right = 0, max(edge[2] for edge in edges)
        
        def feasible(cost: int) -> bool:
            uf = UnionFind(n)
            for u, v, w in edges:
                if w <= cost:
                    uf.union(u, v)
            return uf.count <= k
        
        while left < right:
            mid = (left + right) // 2
            if feasible(mid):
                right = mid
            else:
                left = mid + 1
        return left

Implementation Walkthrough: We first define a Union-Find class with path compression and union by rank. The minCost function sets up a binary search over edge weights. The feasible helper simulates removing edges above a candidate weight. The binary search adjusts the range until we find the minimal feasible cost.

Go Solution

package main

type UnionFind struct {
	parent []int
	rank   []int
	count  int
}

func NewUnionFind(n int) *UnionFind {
	parent := make([]int, n)
	for i := 0; i < n; i++ {
		parent[i] = i
	}
	return &UnionFind{parent, make([]int, n), n}
}

func (uf *UnionFind) Find(x int) int {
	if uf.parent[x] != x {
		uf.parent[x] = uf.Find(uf.parent[x])
	}
	return uf.parent[x]
}

func (uf *UnionFind) Union(x, y int) bool {
	xr, yr := uf.Find(x), uf.Find(y)
	if xr == yr {
		return false
	}
	if uf.rank[xr] < uf.rank[yr] {
		uf.parent[xr] = yr
	} else if uf.rank[xr] > uf.rank[yr] {
		uf.parent[yr] = xr
	} else {
		uf.parent[yr] = xr
		uf.rank[xr]++
	}
	uf.count--
	return true
}

func minCost(n int, edges [][]int, k int) int {
	left, right := 0, 0
	for _, e := range edges {
		if e[2] > right {
			right = e[2]
		}
	}

	feasible := func(cost int) bool {
		uf := NewUnionFind(n)
		for _, e := range edges {
			if e[2] <= cost {
				uf.Union(e[0], e[1])
			}
		}
		return uf.count <= k
	}

	for left < right {
		mid := (left + right) / 2
		if feasible(mid) {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

Go Notes: Go requires explicit slice initialization for the Union-Find arrays. Path compression and rank updates mirror Python logic. We use a closure for feasible to capture edges and n for binary search.

Worked Examples

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

Binary search steps:

left right mid Components ≤ k? Action
0 6 3 3 left = 4
4 6 5 2 right = 5
4 5 4 2 right = 4

Return 4.

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

Binary search