<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 Solution {

private:
    vector<int> parent, size;
    int Find(int node){
        if (parent[node] == node) return node;
        return parent[node] = Find(parent[node]);
    }

    void Union(int node1, int node2){
        int root1 = Find(node1);
        int root2 = Find(node2);
        if (root1 == root2) return;
        if (size[root1] < size[root2]){
            parent[root1] = root2;
            size[root2] += size[root1];
        }
        else{
            parent[root2] = root1;
            size[root1] += size[root2];
        }
    }


public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) return true;
        int MAX = 1e5;
        vector<bool> has(MAX + 1, false);
        for(int& num : nums){
            has[num] = true;
        }

        if (has[1]) return false;
        
        vector<int> sieve(MAX + 1, 0);
        for(int i = 2; i <= MAX; i++){
            if (sieve[i] == 0){
                for(int j = i; j <= MAX; j += i){
                    if (sieve[j] == 0)
                        sieve[j] = i;
                }
            }
        }

        size.resize(2 * MAX + 1, 1);
        parent.resize(2 * MAX + 1);
        for(int i = 0; i < 2 * MAX + 1; i++) parent[i] = i;

        for(auto& num : nums){
            int val = num;
            while(val > 1){
                int spf = sieve[val];
                int root = spf + MAX;
                if (Find(num) != Find(root)) Union(num, root);
                while(val % spf == 0) val /= spf;
            }
        }

        int count = 0;
        for(int i = 2; i <= MAX; i++){
            if (has[i] && Find(i) == i) count++;
        }

        return count == 1;
    }
};
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