How do you solve Largest Component Size by Common Factor on LeetCode?

Largest Component Size by Common Factor (LeetCode #952) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log k) where k is the number of unique prime factors across all numbers. time complexity and O(n) space complexity. Key insight: Understanding prime factorization is crucial for connecting components.

What is the time complexity of Largest Component Size by Common Factor?

The optimal solution for Largest Component Size by Common Factor runs in O(n log k) where k is the number of unique prime factors across all numbers. time and uses O(n) space. This complexity arises from the need to process each number and its prime factors, leading to efficient unions and finds in the Union-Find structure.

What is the brute force solution for Largest Component Size by Common Factor?

The brute force approach for Largest Component Size by Common Factor is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we check every pair of numbers to see if they share a common factor greater than 1. If they do, we connect them in a graph and then find the size of connected components. The optimal Optimal Solution approach improves this to O(n log k) where k is the number of unique prime factors across all numbers..

What algorithm or data structure is used to solve Largest Component Size by Common Factor?

Largest Component Size by Common Factor 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 Largest Component Size by Common Factor?

Explain your thought process clearly while coding. Discuss the trade-offs between different approaches. Practice implementing Union-Find as it is a common pattern in graph problems.

What are common mistakes when solving Largest Component Size by Common Factor?

Forgetting to handle all prime factors of a number. Not using path compression in the Union-Find structure.

#952

Largest Component Size by Common Factor

Hard

ArrayHash TableMathUnion-FindNumber TheoryHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log k) where k is the number of unique prime factors across all numbers.
Space	O(1)	O(n)

💡

Intuition

Time O(n log k) where k is the number of unique prime factors across all numbers.Space O(n)

The optimal approach leverages the Union-Find data structure to efficiently group numbers that share common factors. This allows us to quickly find and union components without explicitly constructing a graph.

⚙️

Algorithm

5 steps

1Step 1: Initialize a Union-Find structure to manage connected components.
2Step 2: For each number, find all its prime factors.
3Step 3: Union all numbers that share the same prime factor.
4Step 4: Count the size of each component using the Union-Find structure.
5Step 5: Return the size of the largest component.

solution.py49 lines

1# Full working Python code
2from collections import defaultdict
3
4class UnionFind:
5    def __init__(self, size):
6        self.parent = list(range(size))
7        self.rank = [1] * size
8
9    def find(self, x):
10        if self.parent[x] != x:
11            self.parent[x] = self.find(self.parent[x])
12        return self.parent[x]
13
14    def union(self, x, y):
15        rootX = self.find(x)
16        rootY = self.find(y)
17        if rootX != rootY:
18            if self.rank[rootX] > self.rank[rootY]:
19                self.parent[rootY] = rootX
20            elif self.rank[rootX] < self.rank[rootY]:
21                self.parent[rootX] = rootY
22            else:
23                self.parent[rootY] = rootX
24                self.rank[rootX] += 1
25
26def largestComponentSize(nums):
27    max_num = max(nums)
28    uf = UnionFind(max_num + 1)
29    prime_to_index = defaultdict(list)
30
31    for num in nums:
32        for factor in range(2, int(num**0.5) + 1):
33            if num % factor == 0:
34                prime_to_index[factor].append(num)
35                while num % factor == 0:
36                    num //= factor
37        if num > 1:
38            prime_to_index[num].append(num)
39
40    for indices in prime_to_index.values():
41        for i in range(1, len(indices)):
42            uf.union(indices[0], indices[i])
43
44    component_size = defaultdict(int)
45    for num in nums:
46        root = uf.find(num)
47        component_size[root] += 1
48
49    return max(component_size.values())

ℹ

Complexity note: This complexity arises from the need to process each number and its prime factors, leading to efficient unions and finds in the Union-Find structure.

1Understanding prime factorization is crucial for connecting components.
2Union-Find is a powerful tool for managing dynamic connectivity.

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