<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 {
public:
    map<int,int> f; //id of factors
    int m = 0; //count of factors

    //edges for numbers & factors
    //N: numbers, F: factors
    vector<int> N[100001], F[100001];

    void add_edge(int id_n, int factor)
    {
        int id_f = f[factor];
        N[id_n].push_back(id_f);
        F[id_f].push_back(id_n);
    }

    void build_graph(int id, vector<int>& nums)
    {
        //check if nums[id] is a prime
        bool isPrime = true;
        
        //find factors
        int k = nums[id], t = sqrt(k);
        for(int i=2; i<=t; i++)
        {
            //add edges to factors
            if(k%i == 0)
            {
                isPrime = false;

                if(f.find(i) == f.end())
                    f[i] = m++;
                
                if(f.find(k/i) == f.end())
                    f[k/i] = m++;
                
                add_edge(id,i);
                add_edge(id,k/i);
            }
        }

        //nums[id] is a prime
        if(isPrime)
        {
            f[k] = m++;
            add_edge(id,k);
        }
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        if(nums[0] == 1 && nums.size() > 1) return false;

        //remove duplicates
        vector<int> num;
        int last = -1;
        for(auto k: nums)
        {
            if(k != last)
            {
                num.push_back(k);
                last = k;
            }
        }

        int n = num.size();
        //build bipartite graph
        for(int i=0; i<n; i++)
            build_graph(i,num);

        //mark if numbers & factors are visited
        bool vis_N[n], vis_F[m];
        for(int i=0; i<n; i++) vis_N[i] = false;
        for(int i=0; i<m; i++) vis_F[i] = false;

        //bipartite graph BFS
        queue<int> q;
        q.push(0);
        while(!q.empty())
        {
            int id_n = q.front();
            q.pop();

            //mark id_n as found
            if(vis_N[id_n]) continue;
            vis_N[id_n] = true;
            
            //edges of id_n
            for(auto id_f: N[id_n])
            {
                //mark id_f as found
                if(vis_F[id_f]) continue;
                vis_F[id_f] = true;
                
                //edges of id_f
                for(auto next_num: F[id_f])
                    q.push(next_num);
            }
        }

        //check if all numbers can be found
        bool possible = true;
        for(int i=0; i<n; i++)
        {
            if(!vis_N[i])
            {
                possible = false;
                break;
            }
        }
        return possible;
    }
};
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