LeetCode 2709: Greatest Common Divisor Traversal Solution

Master LeetCode problem 2709 (Greatest Common Divisor Traversal), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2709. Greatest Common Divisor Traversal

Hard

Array Math Union Find Number Theory

Problem Explanation

Explanation

To solve this problem, we can construct a graph where each index in the array nums represents a node and there is a directed edge between nodes i and j if gcd(nums[i], nums[j]) > 1. We then need to check if this graph is strongly connected, meaning that there is a path from every node to every other node.

We can use Depth First Search (DFS) to traverse the graph and check if all nodes can be visited from any starting node. If all nodes can be visited, then the answer is true; otherwise, it is false.

Time Complexity: O(n * sqrt(m)), where n is the number of elements in nums and m is the maximum element in nums. Space Complexity: O(n)

Solution Code

import java.util.*;

class Solution {
    public boolean canTraverse(int[] nums) {
        int n = nums.length;
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int i = 0; i < n; i++) {
            graph.put(i, new ArrayList<>());
            for (int j = 0; j < n; j++) {
                if (i != j && gcd(nums[i], nums[j]) > 1) {
                    graph.get(i).add(j);
                }
            }
        }
        
        boolean[] visited = new boolean[n];
        
        for (int i = 0; i < n; i++) {
            if (dfs(i, graph, visited)) {
                return true;
            }
        }
        
        return false;
    }
    
    private boolean dfs(int node, Map<Integer, List<Integer>> graph, boolean[] visited) {
        if (visited[node]) {
            return false;
        }
        
        visited[node] = true;
        
        for (int neighbor : graph.get(node)) {
            if (!visited[neighbor] && dfs(neighbor, graph, visited)) {
                return true;
            }
        }
        
        visited[node] = false;
        
        return false;
    }
    
    private int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

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