LeetCode #3613 — MEDIUM

Minimize Maximum Component Cost

Move from brute-force thinking to an efficient approach using binary search strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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] = [ui, vi, wi] denotes an undirected edge between node ui and node vi with weight wi, 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Binary Search · Union-Find

Example 1

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

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

Sort the <code>edges</code> and do binary search on the candidate maximum weight
Use <code>DFS</code> or <code>DSU</code> to count the number of connected components when keeping only edges with weight <= mid

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3613: Minimize Maximum Component Cost
class Solution {
    private int[] p;

    public int minCost(int n, int[][] edges, int k) {
        if (k == n) {
            return 0;
        }
        p = new int[n];
        Arrays.setAll(p, i -> i);
        Arrays.sort(edges, Comparator.comparingInt(a -> a[2]));
        int cnt = n;
        for (var e : edges) {
            int u = e[0], v = e[1], w = e[2];
            int pu = find(u), pv = find(v);
            if (pu != pv) {
                p[pu] = pv;
                if (--cnt <= k) {
                    return w;
                }
            }
        }
        return 0;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

// Accepted solution for LeetCode #3613: Minimize Maximum Component Cost
func minCost(n int, edges [][]int, k int) int {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}

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

	if k == n {
		return 0
	}

	slices.SortFunc(edges, func(a, b []int) int {
		return a[2] - b[2]
	})

	cnt := n
	for _, e := range edges {
		u, v, w := e[0], e[1], e[2]
		pu, pv := find(u), find(v)
		if pu != pv {
			p[pu] = pv
			if cnt--; cnt <= k {
				return w
			}
		}
	}

	return 0
}

# Accepted solution for LeetCode #3613: Minimize Maximum Component Cost
class Solution:
    def minCost(self, n: int, edges: List[List[int]], k: int) -> int:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        if k == n:
            return 0
        edges.sort(key=lambda x: x[2])
        cnt = n
        p = list(range(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
        return 0

// Accepted solution for LeetCode #3613: Minimize Maximum Component Cost
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3613: Minimize Maximum Component Cost
// class Solution {
//     private int[] p;
// 
//     public int minCost(int n, int[][] edges, int k) {
//         if (k == n) {
//             return 0;
//         }
//         p = new int[n];
//         Arrays.setAll(p, i -> i);
//         Arrays.sort(edges, Comparator.comparingInt(a -> a[2]));
//         int cnt = n;
//         for (var e : edges) {
//             int u = e[0], v = e[1], w = e[2];
//             int pu = find(u), pv = find(v);
//             if (pu != pv) {
//                 p[pu] = pv;
//                 if (--cnt <= k) {
//                     return w;
//                 }
//             }
//         }
//         return 0;
//     }
// 
//     private int find(int x) {
//         if (p[x] != x) {
//             p[x] = find(p[x]);
//         }
//         return p[x];
//     }
// }

// Accepted solution for LeetCode #3613: Minimize Maximum Component Cost
function minCost(n: number, edges: number[][], k: number): number {
    const p: number[] = Array.from({ length: n }, (_, i) => i);
    const find = (x: number): number => {
        if (p[x] !== x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };

    if (k === n) {
        return 0;
    }

    edges.sort((a, b) => a[2] - b[2]);
    let cnt = n;
    for (const [u, v, w] of edges) {
        const pu = find(u),
            pv = find(v);
        if (pu !== pv) {
            p[pu] = pv;
            if (--cnt <= k) {
                return w;
            }
        }
    }
    return 0;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.