2709. Greatest Common Divisor Traversal

Problem Link

Description

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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.

Your task is to determine if for every pair of indices i and j in nums, where i < j, there exists a sequence of traversals that can take us from i to j.

Return true if it is possible to traverse between all such pairs of indices, or false otherwise.

 

Example 1:

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.

Example 2:

Input: nums = [3,9,5]
Output: false
Explanation: No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.

Example 3:

Input: nums = [4,3,12,8]
Output: true
Explanation: There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105

Solution

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Python3

class DSU:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.rank = [0 for i in range(n)]
 
    def find(self, x):
        if self.parent[x] == x:
            return self.parent[x]
        self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
 
    def union(self, u, v):
        pu = self.find(u)
        pv = self.find(v)
 
        if self.rank[pu] < self.rank[v]:
            pu, pv = pv, pu
 
        self.parent[pv] = pu
        self.rank[pu] += self.rank[pv]
 
class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        N = len(nums)
        uf = DSU(N)
        factors = defaultdict(list)
 
        for i, x in enumerate(nums):
            for factor in range(2, int(math.sqrt(x)) + 1):
                if x % factor == 0:
                    factors[factor].append(i)
 
                    while x % factor == 0:
                        x //= factor
                
            if x > 1:
                factors[x].append(i)
        
        for A in factors.values():
            for a, b in zip(A, A[1:]):
                uf.union(a, b)
        
        p = uf.find(0)
        for i in range(N):
            if uf.find(i) != p: return False
        
        return True