#3608

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Minimum Time for K Connected Components

Find the minimum time to remove edges such that a graph with n nodes has at least k connected components.

Problem Statement

You are given an integer n representing the number of nodes in an undirected graph, with edges given in a 2D array. Each edge is defined by three values: ui, vi, and timei, indicating an undirected edge between nodes ui and vi that can be removed at time timei.

You are also provided with an integer k, and your goal is to find the minimum time t such that after removing all edges with time <= t, the graph will be split into at least k connected components.

Examples

Example 1

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

Output: 3

Example 2

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

Output: 4

Example 3

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

Output: 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.

Solution Approach

Binary Search over Time

The problem can be approached by performing a binary search over the possible edge removal times. The smallest time t that results in the graph splitting into at least k connected components is the answer.

Union-Find for Component Tracking

Union-Find (Disjoint Set Union) will be used to efficiently track connected components as edges are removed. After each binary search step, union-find helps check if the graph is split into the desired number of components.

Edge Removal Simulation

For each candidate time t found by the binary search, simulate the removal of all edges with time <= t. Union-find operations will help check if the number of connected components is >= k at that point.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is dominated by the binary search, which runs in O(log(max_time)), and the union-find operations for each step, resulting in a total time complexity of O(log(max_time) * n). The space complexity is O(n) due to the union-find data structure.

What Interviewers Usually Probe

The candidate must demonstrate a clear understanding of binary search principles and how they apply to this problem.
The candidate should effectively use union-find to manage connected components during the binary search.
A solid solution will include reasoning on time complexity and justify the approach with optimal space utilization.

Common Pitfalls or Variants

Common pitfalls

Candidates might attempt to use brute force methods instead of binary search, leading to inefficient solutions.
Misunderstanding the problem constraints and removing edges prematurely may cause incorrect results.
Failing to optimize union-find operations with path compression and union by rank could lead to slower performance.

Follow-up variants

What if the graph starts already with k connected components?
How can this approach be modified for directed graphs?
What happens if there are more edges with the same time value?

FAQ

What is the binary search approach for the 'Minimum Time for K Connected Components' problem?

The binary search approach involves testing different removal times and checking if the graph splits into k connected components using union-find.

How does union-find help in solving the problem?

Union-find helps track the number of connected components as edges are removed, enabling efficient checks of graph connectivity at each binary search step.

What is the minimum time for the graph to have 3 components in this problem?

The minimum time is determined by performing binary search to find the smallest time t where removing all edges with time <= t results in at least 3 connected components.

Why does binary search work for this problem?

Binary search works by narrowing down the valid answer space for time t, testing each time to find the smallest t that creates at least k components.

What are the main challenges in solving the 'Minimum Time for K Connected Components' problem?

The main challenges are implementing an efficient binary search and optimizing union-find operations to handle large input sizes and edge cases.

terminal

Solution

Solution 1: Union-Find

We can sort the edges by time in ascending order, then starting from the edge with the largest time, add edges to the graph one by one, while using a union-find data structure to maintain the number of connected components in the current graph. When the number of connected components is less than $k$, the current time is the minimum time we are looking for.

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class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))
        self.size = [1] * n

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, a, b):
        pa, pb = self.find(a), self.find(b)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        return True


class Solution:
    def minTime(self, n: int, edges: List[List[int]], k: int) -> int:
        edges.sort(key=lambda x: x[2])
        uf = UnionFind(n)
        cnt = n
        for u, v, t in edges[::-1]:
            if uf.union(u, v):
                cnt -= 1
                if cnt < k:
                    return t
        return 0

Continue Practicing

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