<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 UnionFind {
private:
    int n, *parent, *rank;
public:
    int set_size;
    UnionFind(){}
    UnionFind(int a) { //a: set size, {1,2...n}
        n=set_size=a;
        parent=new int[n+1];
        rank=new int[n+1];
        for (int i=0;i<n;i++) parent[i]=i, rank[i]=1;
    }

    int find(int x) { //O(A(n))
        if (x!=parent[x]) return parent[x]=find(parent[x]); //path-compression
        return x;
    }

    bool merge(int x, int y) { //O(A(n))
        int xroot=find(x), yroot=find(y);
        if (xroot!=yroot) { //rank-compression
            if (rank[xroot]>=rank[yroot]) {
                parent[yroot]=xroot;
                rank[xroot]+=rank[yroot];
            } else {
                parent[xroot]=yroot;
                rank[yroot]+=rank[xroot];
            }
            set_size--;
            return 1;
        }
        return 0;
    }

    void reset() {
        set_size=n;
        for (int i=0;i<n;i++) parent[i]=i, rank[i]=1;  
    }
};

unordered_set<int> get_divisors(int n) { //O(sqrt(n))
    unordered_set<int> ans;
    int _=sqrt(n);
    for (int i=2;i<=_;i++)
        while (!(n%i)) {
            ans.insert(i);
            n/=i;
        }
    if (n>1) ans.insert(n);
    return ans;
}
class Solution {
public:
bool canTraverseAllPairs(vector<int>& nums) {
    unordered_map<int, vector<int>> mp;
    for (int i=0;i<nums.size();i++) //O(nums.length*sqrt(nums[i]))
        for (auto &_:get_divisors(nums[i]))
            mp[_].push_back(i);

    vector<vector<int>> g;g.resize(nums.size());
    UnionFind uf(nums.size());
    for (auto &[_,arr]:mp) 
        for (int i=1;i<arr.size();i++) uf.merge(arr[i-1], arr[i]);
    return uf.set_size==1;
}
};
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