How do you solve Greatest Common Divisor Traversal on LeetCode?

Greatest Common Divisor Traversal (LeetCode #2709) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max(nums))) time complexity and O(n) space complexity. Key insight: Using prime factors helps reduce the number of connections we need to check.

What is the time complexity of Greatest Common Divisor Traversal?

The optimal solution for Greatest Common Divisor Traversal runs in O(n log(max(nums))) time and uses O(n) space. The time complexity is O(n log(max(nums))) due to the factorization process for each number. The space complexity is O(n) for the union-find structure.

What is the brute force solution for Greatest Common Divisor Traversal?

The brute force approach for Greatest Common Divisor Traversal is Brute Force, with O(n²) time complexity and O(n) space complexity. We can check all pairs of indices to see if there's a path between them based on the GCD condition. This is straightforward but inefficient as it checks every possible pair. The optimal Optimal Solution approach improves this to O(n log(max(nums))).

What algorithm or data structure is used to solve Greatest Common Divisor Traversal?

Greatest Common Divisor Traversal uses the following concepts: Array, Math, Union-Find, Number Theory. The recommended patterns to study are: Union-Find, Graph Traversal.

What are the interview tips for Greatest Common Divisor Traversal?

Think about data structures that can help with connectivity, like Union-Find. Practice factorization techniques, as they can simplify problems involving GCD. Always consider edge cases, such as arrays with all prime numbers.

What are common mistakes when solving Greatest Common Divisor Traversal?

Not optimizing the GCD checks, leading to timeouts. Forgetting to handle cases where numbers are prime.

#2709

Greatest Common Divisor Traversal

Hard

ArrayMathUnion-FindNumber TheoryUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log(max(nums)))Space O(n)

Instead of checking all pairs, we can use a union-find (disjoint set) structure to efficiently group indices based on their GCD connections. This allows us to determine if all indices are connected in one pass.

⚙️

Algorithm

3 steps

1Step 1: Create a union-find structure to manage connected components.
2Step 2: For each number, find its prime factors and union all indices that share a prime factor.
3Step 3: After processing all numbers, check if all indices belong to the same 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, u):
10        if self.parent[u] != u:
11            self.parent[u] = self.find(self.parent[u])
12        return self.parent[u]
13
14    def union(self, u, v):
15        rootU = self.find(u)
16        rootV = self.find(v)
17        if rootU != rootV:
18            if self.rank[rootU] > self.rank[rootV]:
19                self.parent[rootV] = rootU
20            elif self.rank[rootU] < self.rank[rootV]:
21                self.parent[rootU] = rootV
22            else:
23                self.parent[rootV] = rootU
24                self.rank[rootU] += 1
25
26def canTraverse(nums):
27    n = len(nums)
28    uf = UnionFind(n)
29    prime_to_index = defaultdict(list)
30
31    for i in range(n):
32        num = nums[i]
33        for factor in range(2, int(num**0.5) + 1):
34            if num % factor == 0:
35                prime_to_index[factor].append(i)
36                while num % factor == 0:
37                    num //= factor
38        if num > 1:
39            prime_to_index[num].append(i)
40
41    for indices in prime_to_index.values():
42        for j in range(1, len(indices)):
43            uf.union(indices[0], indices[j])
44
45    root = uf.find(0)
46    return all(uf.find(i) == root for i in range(n))
47
48# Example usage
49print(canTraverse([2, 3, 6]))  # Output: True

ℹ

Complexity note: The time complexity is O(n log(max(nums))) due to the factorization process for each number. The space complexity is O(n) for the union-find structure.

1Using prime factors helps reduce the number of connections we need to check.
2Union-Find is effective for dynamic connectivity problems.

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