How do you solve Count Connected Components in LCM Graph on LeetCode?

Count Connected Components in LCM Graph (LeetCode #3378) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Use Union-Find for efficient grouping.

What is the time complexity of Count Connected Components in LCM Graph?

The optimal solution for Count Connected Components in LCM Graph runs in O(n log n) time and uses O(n) space. The complexity is linearithmic due to the union-find operations and multiple connections.

What is the brute force solution for Count Connected Components in LCM Graph?

The brute force approach for Count Connected Components in LCM Graph is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every pair of nodes to see if they are connected based on the LCM condition. This method is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Count Connected Components in LCM Graph?

Count Connected Components in LCM Graph uses the following concepts: Array, Hash Table, Math, Union-Find, Number Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Connected Components in LCM Graph?

Explain your thought process clearly. Discuss trade-offs between brute force and optimal solutions. Practice similar problems to build confidence.

What are common mistakes when solving Count Connected Components in LCM Graph?

Forgetting to check all multiples. Not using union-find optimally.

#3378

Count Connected Components in LCM Graph

Hard

ArrayHash TableMathUnion-FindNumber TheoryHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

Use Union-Find to efficiently group numbers by connecting multiples that are less than the threshold, reducing unnecessary comparisons.

⚙️

Algorithm

3 steps

1Step 1: Initialize a Union-Find structure to manage connected components.
2Step 2: For each number, connect it to all its multiples that are less than or equal to the threshold.
3Step 3: Count the distinct roots in the Union-Find structure to determine the number of connected components.

solution.py20 lines

1class UnionFind:
2    def __init__(self, size):
3        self.parent = list(range(size))
4    def find(self, x):
5        if self.parent[x] != x:
6            self.parent[x] = self.find(self.parent[x])
7        return self.parent[x]
8    def union(self, x, y):
9        rootX = self.find(x)
10        rootY = self.find(y)
11        if rootX != rootY:
12            self.parent[rootY] = rootX
13
14def countComponents(nums, threshold):
15    uf = UnionFind(len(nums))
16    for i in range(len(nums)):
17        for j in range(2, threshold // nums[i] + 1):
18            if j * nums[i] in nums:
19                uf.union(i, nums.index(j * nums[i]))
20    return len(set(uf.find(i) for i in range(len(nums))))

ℹ

Complexity note: The complexity is linearithmic due to the union-find operations and multiple connections.

1Use Union-Find for efficient grouping.
2LCM properties can reduce unnecessary checks.

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