How do you solve Minimum Time for K Connected Components on LeetCode?

Minimum Time for K Connected Components (LeetCode #3608) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m log m + m α(n)) time complexity and O(n) space complexity. Key insight: Binary search reduces the number of checks needed.

What is the time complexity of Minimum Time for K Connected Components?

The optimal solution for Minimum Time for K Connected Components runs in O(m log m + m α(n)) time and uses O(n) space. Sorting edges takes O(m log m), and Union-Find operations take O(m α(n)), where α is the inverse Ackermann function.

What is the brute force solution for Minimum Time for K Connected Components?

The brute force approach for Minimum Time for K Connected Components is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible time to see if removing edges results in at least k connected components. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(m log m + m α(n)).

What algorithm or data structure is used to solve Minimum Time for K Connected Components?

Minimum Time for K Connected Components uses the following concepts: Binary Search, Union-Find, Graph Theory, Sorting. The recommended patterns to study are: Binary Search, Union-Find.

What are the interview tips for Minimum Time for K Connected Components?

Explain your thought process clearly. Discuss edge cases and how you would handle them. Practice similar problems to build confidence.

What are common mistakes when solving Minimum Time for K Connected Components?

Not sorting edges before processing. Incorrectly implementing the Union-Find structure.

#3608

Minimum Time for K Connected Components

Medium

Binary SearchUnion-FindGraph TheorySortingBinary SearchUnion-Find

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m log m + m α(n))
Space	O(1)	O(n)

💡

Intuition

Time O(m log m + m α(n))Space O(n)

Use binary search on time to efficiently find the minimum time t where the graph has at least k components after edge removals.

⚙️

Algorithm

3 steps

1Step 1: Sort edges by time.
2Step 2: Use binary search on time from 0 to max edge time.
3Step 3: For mid time, use Union-Find to count components and adjust search range based on the count.

solution.py14 lines

1def minTime(n, edges, k):
2    edges.sort(key=lambda x: x[2])
3    left, right = 0, edges[-1][2]
4    while left < right:
5        mid = (left + right) // 2
6        uf = UnionFind(n)
7        for u, v, time in edges:
8            if time <= mid:
9                uf.union(u, v)
10        if uf.count() >= k:
11            right = mid
12        else:
13            left = mid + 1
14    return left

ℹ

Complexity note: Sorting edges takes O(m log m), and Union-Find operations take O(m α(n)), where α is the inverse Ackermann function.

1Binary search reduces the number of checks needed.
2Union-Find efficiently counts connected components.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.