LeetCode 3613 - Minimize Maximum Component Cost

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

Solution

Problem Understanding

The problem asks us to minimize the maximum cost among components in a connected undirected graph after removing edges. Each edge has a weight, and the cost of a connected component is the largest weight of any edge in that component. You can remove edges freely, but the graph must end up with at most k connected components.

The input consists of n nodes, a list of weighted edges edges, and an integer k. The expected output is the minimum possible value of the largest component cost after removing edges. The constraints tell us the graph can be quite large (n up to 50,000 nodes and edges.length up to 100,000), so brute-force checking all combinations of edge removals is infeasible. Edge cases to note include graphs with a single component (k = 1) where no edges can be removed, and graphs with many nodes but very few edges.

The problem guarantees the graph is connected, so there will always be at least one way to keep it connected. Also, weights are positive integers, which avoids issues with zero or negative weights when calculating component costs.

Approaches

The brute-force approach would be to try every combination of edges to remove and compute the resulting component costs. For each configuration, we calculate the maximum edge weight in each connected component and take the largest among them. This is correct but impractical because the number of ways to remove edges grows exponentially with the number of edges. This would be O(2^E) where E is the number of edges, far too large for the given constraints.

The key insight is that we do not need to try every combination of edge removals. Since the cost of a component is determined by its maximum edge weight, we can focus on whether it is possible to split the graph into k components with all edges having weight ≤ some candidate mid. This suggests a binary search over possible maximum edge weights. For a given weight mid, we can use Union-Find to check how many connected components remain if we only keep edges with weight ≤ mid. If the number of components is ≤ k, the weight is feasible. Otherwise, we need a larger weight.

This transforms the problem into a binary search problem combined with a Union-Find check.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^E * (V + E))	O(V)	Try all subsets of edges, calculate components each time; infeasible
Optimal	O(E log W + α(V) * E)	O(V)	Binary search on max weight W and Union-Find to count components; feasible for constraints

Algorithm Walkthrough

Sort Edges by Weight: Start by noting the minimum and maximum possible edge weights. This defines the search space for binary search.
Binary Search Setup: Initialize left as the smallest weight and right as the largest weight. We aim to find the smallest weight mid such that the graph can be partitioned into ≤ k components with all edges ≤ mid.
Union-Find Helper Function: Implement a function that takes a weight max_weight and counts the number of connected components if only edges with weight ≤ max_weight are considered. Use Union-Find with path compression and union by size for efficiency.
Binary Search Loop: While left < right, compute mid = (left + right) // 2. Check the number of components using the Union-Find helper. If the number of components is ≤ k, reduce right = mid to search for a smaller feasible max weight. Otherwise, increase left = mid + 1.
Return Result: Once left == right, the smallest feasible maximum edge weight is found, which is the answer.

Why it works: The algorithm maintains the invariant that left and right always bracket the smallest feasible maximum component cost. The Union-Find check guarantees the component count is correctly determined for a given threshold, ensuring correctness.

Python Solution

from typing import List

class Solution:
    def minCost(self, n: int, edges: List[List[int]], k: int) -> int:
        parent = list(range(n))
        size = [1] * n

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

        def union(x: int, y: int) -> bool:
            xr, yr = find(x), find(y)
            if xr == yr:
                return False
            if size[xr] < size[yr]:
                xr, yr = yr, xr
            parent[yr] = xr
            size[xr] += size[yr]
            return True

        def count_components(max_weight: int) -> int:
            for i in range(n):
                parent[i] = i
                size[i] = 1
            components = n
            for u, v, w in edges:
                if w <= max_weight and union(u, v):
                    components -= 1
            return components

        left, right = 1, max(w for _, _, w in edges)
        while left < right:
            mid = (left + right) // 2
            if count_components(mid) <= k:
                right = mid
            else:
                left = mid + 1
        return left

Explanation: We initialize Union-Find arrays parent and size. The find function implements path compression, and union merges sets. The count_components function rebuilds Union-Find for each candidate mid and counts components. Binary search efficiently narrows down the minimum feasible maximum weight.

Go Solution

func minCost(n int, edges [][]int, k int) int {
    parent := make([]int, n)
    size := make([]int, n)
    for i := 0; i < n; i++ {
        parent[i] = i
        size[i] = 1
    }

    var find func(int) int
    find = func(x int) int {
        if parent[x] != x {
            parent[x] = find(parent[x])
        }
        return parent[x]
    }

    union := func(x, y int) bool {
        xr, yr := find(x), find(y)
        if xr == yr {
            return false
        }
        if size[xr] < size[yr] {
            xr, yr = yr, xr
        }
        parent[yr] = xr
        size[xr] += size[yr]
        return true
    }

    countComponents := func(maxWeight int) int {
        for i := 0; i < n; i++ {
            parent[i] = i
            size[i] = 1
        }
        components := n
        for _, e := range edges {
            u, v, w := e[0], e[1], e[2]
            if w <= maxWeight && union(u, v) {
                components--
            }
        }
        return components
    }

    left, right := 1, 0
    for _, e := range edges {
        if e[2] > right {
            right = e[2]
        }
    }

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

Go-specific notes: Arrays are initialized manually. Function closures are used for find and countComponents. Slice semantics are used for edges. Otherwise, the logic mirrors the Python implementation.

Worked Examples

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

Binary search range: 1 to 6.

Check mid = 3: components = 3 > 2 → increase left.

Check mid = 5: components = 2 ≤ 2 → decrease right.

Result converges to 4, which matches the expected answer.

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

Only one component allowed. Maximum edge weight cannot be reduced. Binary search confirms 5 is minimal.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(E log W * α(V))	Binary search on edge weights (log W) and counting components via Union-Find (α(V) amortized)
Space	O(V)	Union-Find arrays `parent` and `size`

The binary search iterates at most log W times, and each iteration traverses all edges while performing near-constant time union-find operations.

Test Cases

# Provided examples
assert Solution().minCost(5, [[0,1,4],[1,2,3],[1,3,2],[3,4,6]], 2) == 4  # example 1
assert Solution().minCost(4, [[0,1,5],[1,2,5],[2,3,5]], 1) == 5  # example 2

# Edge cases
assert Solution().minCost(1, [], 1) == 0  # single node, no edges
assert