LeetCode #2709 — HARD

Greatest Common Divisor Traversal

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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.

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
Patterns Used
🔗Union-Find→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Union-Find

Example 1

[2,3,6]

Example 2

[3,9,5]

Example 3

[4,3,12,8]

Related Problems

  • Graph Connectivity With Threshold (graph-connectivity-with-threshold)
Step 02

Core Insight

What unlocks the optimal approach

  • Create a (prime) factor-numbers list for all the indices.
  • Add an edge between the neighbors of the (prime) factor-numbers list. The order of the numbers doesn’t matter. We only need edges between 2 neighbors instead of edges for all pairs.
  • The problem is now similar to checking if all the numbers (nodes of the graph) are in the same connected component.
  • Any algorithm (i.e., BFS, DFS, or Union-Find Set) should work to find or check connected components
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2709: Greatest Common Divisor Traversal
class UnionFind {
    private int[] p;
    private int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    public boolean union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }
}

class Solution {
    private static final int MX = 100010;
    private static final List<Integer>[] P = new List[MX];

    static {
        Arrays.setAll(P, k -> new ArrayList<>());
        for (int x = 1; x < MX; ++x) {
            int v = x;
            int i = 2;
            while (i <= v / i) {
                if (v % i == 0) {
                    P[x].add(i);
                    while (v % i == 0) {
                        v /= i;
                    }
                }
                ++i;
            }
            if (v > 1) {
                P[x].add(v);
            }
        }
    }

    public boolean canTraverseAllPairs(int[] nums) {
        int m = Arrays.stream(nums).max().getAsInt();
        int n = nums.length;
        UnionFind uf = new UnionFind(n + m + 1);
        for (int i = 0; i < n; ++i) {
            for (int j : P[nums[i]]) {
                uf.union(i, j + n);
            }
        }
        Set<Integer> s = new HashSet<>();
        for (int i = 0; i < n; ++i) {
            s.add(uf.find(i));
        }
        return s.size() == 1;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(α(n))
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND
O(α(n)) time
O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Related Problems
#952Largest Component Size by Common Factorhard→#1569Number of Ways to Reorder Array to Get Same BSThard→#1627Graph Connectivity With Thresholdhard→#1998GCD Sort of an Arrayhard→#3235Check if the Rectangle Corner Is Reachablehard→