<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 {
    struct DSU {
        vector<int> par, rank;
        int count = 0;
        DSU(int n)
        {
            par.resize(n);
            iota(par.begin(), par.end(), 0);
            rank.assign(n, 1);
            count = n;
        }
        int find(int x)
        {
            if (par[x] != x)
            {
                par[x] = find(par[x]);
            }

            return par[x];
        }
        bool doUnion(int n1, int n2)
        {
            int p1 = find(n1);
            int p2 = find(n2);

            if (p1 == p2)
                return false;
            
            if (rank[p1] >= rank[p2])
            {
                rank[p1] += rank[p2];
                par[p2] = p1;
            }
            else
            {
                rank[p2] += rank[p1];
                par[p1] = p2;
            }

            count -= 1;
            return true;
        }
    };
public:
    set<int> primeFactors(int num) 
    {
        set<int> factors;
        
        while (num % 2 == 0) 
        {
            factors.insert(2);
            num = num / 2;
        }
        
        for (int i = 3; i <= sqrt(num); i += 2) 
        {
            while (num % i == 0) 
            {
                factors.insert(i);
                num = num / i;
            }
        }
        
        if (num > 2)
            factors.insert(num);
        
        return factors;
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        DSU dsu(nums.size());
        unordered_map<int, int> factor_index;

        for (int i = 0; i < nums.size(); i++)
        {
            set<int> factors = primeFactors(nums[i]);
            for (auto it: factors)
            {
                if (factor_index.count(it))
                    dsu.doUnion(i, factor_index[it]);
                else
                    factor_index[it] = i;
            }
        }
        
        return dsu.count == 1;

    }
};
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