LeetCode #3608 — MEDIUM

Minimum Time for K Connected Components

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

Solve on LeetCode
The Problem

Problem Statement

You are given an integer n and an undirected graph with n nodes labeled from 0 to n - 1. This is represented by a 2D array edges, where edges[i] = [ui, vi, timei] indicates an undirected edge between nodes ui and vi that can be removed at timei.

You are also given an integer k.

Initially, the graph may be connected or disconnected. Your task is to find the minimum time t such that after removing all edges with time <= t, the graph contains at least k connected components.

Return the minimum time t.

A connected component is a subgraph of a graph in which there exists a path between any two vertices, and no vertex of the subgraph shares an edge with a vertex outside of the subgraph.

Example 1:

Input: n = 2, edges = [[0,1,3]], k = 2

Output: 3

Explanation:

  • Initially, there is one connected component {0, 1}.
  • At time = 1 or 2, the graph remains unchanged.
  • At time = 3, edge [0, 1] is removed, resulting in k = 2 connected components {0}, {1}. Thus, the answer is 3.

Example 2:

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

Output: 4

Explanation:

  • Initially, there is one connected component {0, 1, 2}.
  • At time = 2, edge [0, 1] is removed, resulting in two connected components {0}, {1, 2}.
  • At time = 4, edge [1, 2] is removed, resulting in k = 3 connected components {0}, {1}, {2}. Thus, the answer is 4.

Example 3:

Input: n = 3, edges = [[0,2,5]], k = 2

Output: 0

Explanation:

  • Since there are already k = 2 disconnected components {1}, {0, 2}, no edge removal is needed. Thus, the answer is 0.

Constraints:

  • 1 <= n <= 105
  • 0 <= edges.length <= 105
  • edges[i] = [ui, vi, timei]
  • 0 <= ui, vi < n
  • ui != vi
  • 1 <= timei <= 109
  • 1 <= k <= n
  • There are no duplicate edges.
Patterns Used
🔍Modified Binary Search→🔗Union-Find→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an integer n and an undirected graph with n nodes labeled from 0 to n - 1. This is represented by a 2D array edges, where edges[i] = [ui, vi, timei] indicates an undirected edge between nodes ui and vi that can be removed at timei. You are also given an integer k. Initially, the graph may be connected or disconnected. Your task is to find the minimum time t such that after removing all edges with time <= t, the graph contains at least k connected components. Return the minimum time t. A connected component is a subgraph of a graph in which there exists a path between any two vertices, and no vertex of the subgraph shares an edge with a vertex outside of the subgraph.

Baseline thinking

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

Pattern signal: Binary Search · Union-Find

Example 1

2
[[0,1,3]]
2

Example 2

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

Example 3

3
[[0,2,5]]
2
Step 02

Core Insight

What unlocks the optimal approach

  • Binary-search the smallest <code>t</code> such that, after removing all edges with <code>time <= t</code>, the graph splits into >= <code>k</code> connected components.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3608: Minimum Time for K Connected Components
class UnionFind {
    private int[] p;
    private int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; i++) {
            p[i] = i;
            size[i] = 1;
        }
    }

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

    public boolean union(int a, int b) {
        int pa = find(a);
        int pb = find(b);
        if (pa == pb) {
            return false;
        }

        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }
}

class Solution {
    public int minTime(int n, int[][] edges, int k) {
        Arrays.sort(edges, (a, b) -> Integer.compare(a[2], b[2]));

        UnionFind uf = new UnionFind(n);
        int cnt = n;

        for (int i = edges.length - 1; i >= 0; i--) {
            int u = edges[i][0];
            int v = edges[i][1];
            int t = edges[i][2];

            if (uf.union(u, v)) {
                if (--cnt < k) {
                    return t;
                }
            }
        }
        return 0;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × \alpha(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.

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