<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>, and you are allowed to <strong>traverse</strong> between its indices. You can traverse between index <code>i</code> and index <code>j</code>, <code>i != j</code>, if and only if <code>gcd(nums[i], nums[j]) &gt; 1</code>, where <code>gcd</code> is the <strong>greatest common divisor</strong>.</p>

<p>Your task is to determine if for <strong>every pair</strong> of indices <code>i</code> and <code>j</code> in nums, where <code>i &lt; j</code>, there exists a <strong>sequence of traversals</strong> that can take us from <code>i</code> to <code>j</code>.</p>

<p>Return <code>true</code><em> if it is possible to traverse between all such pairs of indices,</em><em> or </em><code>false</code><em> otherwise.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> 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 -&gt; 2 -&gt; 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 &gt; 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 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 &gt; 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,3,12,8]
<strong>Output:</strong> true
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Graph {
private:
    int n;
    vector<vector<int>> edges;

    void traverse(int x, vector<bool>& visited) {
        visited[x] = true;
        for (int y: edges[x]) {
            if (!visited[y]) {
                traverse(y, visited);
            }
        }
    }

public:
    Graph(int n = 0) : n(n) {
        edges.resize(n);
    }
    
    void addEdge(int x, int y) {
        edges[x].push_back(y);
        edges[y].push_back(x);
    }

    bool isConnected() {
        vector<bool> visited(n);
        traverse(0, visited);
        return count(visited.begin(), visited.end(), true) == n;
    }
};

class Solution {
private:
    vector<int> getPrimeFactors(int x) {
        vector<int> primeFactors;
        for (int i = 2; i * i <= x; i++) {
            if (x % i == 0) {
                primeFactors.push_back(i);
                while (x % i == 0) {
                    x /= i;
                }
            }
        }
        if (x != 1) {
            primeFactors.push_back(x);
        }
        return primeFactors;
    }

public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) {
            return true;
        }
        Graph g(n);
        unordered_map<int, int> seen;
        for (int i = 0; i < n; i++) {
            if (nums[i] == 1) {
                return false;
            }
            vector<int> primeFactors = getPrimeFactors(nums[i]);
            for (int prime: primeFactors) {
                if (seen.find(prime) != seen.end()) {
                    g.addEdge(i, seen[prime]);
                } else {
                    seen[prime] = i;
                }
            }
        }
        return g.isConnected();
    }
};
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