Leetcode #2709: Greatest Common Divisor Traversal
In this guide, we solve Leetcode #2709 Greatest Common Divisor Traversal in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed integer array nums, and you are allowed to traverse between its indices. You can traverse between index i and index j, i != j, if and only if gcd(nums[i], nums[j]) > 1, where gcd is the greatest common divisor.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Union Find, Array, Math, Number Theory
Intuition
We need to merge components and check connectivity efficiently.
Union-Find supports near-constant-time merges and finds.
Approach
Initialize each node as its own parent and union pairs as you scan.
Use path compression to keep operations fast.
Steps:
- Initialize parent arrays.
- Union related nodes.
- Use find to check connectivity.
Example
Input: nums = [2,3,6]
Output: true
Explanation: In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).
To go from index 0 to index 1, we can use the sequence of traversals 0 -> 2 -> 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 > 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 > 1.
Python Solution
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
mx = 100010
p = defaultdict(list)
for x in range(1, mx + 1):
v = x
i = 2
while i <= v // i:
if v % i == 0:
p[x].append(i)
while v % i == 0:
v //= i
i += 1
if v > 1:
p[x].append(v)
class Solution:
def canTraverseAllPairs(self, nums: List[int]) -> bool:
n = len(nums)
m = max(nums)
uf = UnionFind(n + m + 1)
for i, x in enumerate(nums):
for j in p[x]:
uf.union(i, j + n)
return len(set(uf.find(i) for i in range(n))) == 1
Complexity
The time complexity is Near O(n) (amortized). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.