<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:
    vector<int> adj[100005];
    int vis[100005];
    void dfs(int u , int p){
        vis[u] = 1;
        for(int v : adj[u]){
            if(v == p)continue;
            if(vis[v] == 0)dfs(v , u);
        }
    }

    // linear sieve
    int lpf[100005]; // lpf[x] : min prime factor of x
    vector<int> prime;
    void lin(){
        for(int i=2;i<100005;i++){
            if(lpf[i] == 0){
                // i is prime
                lpf[i] = i; 
                prime.push_back(i);
                // isPrime[i] = 1;
            }
            for(int j=0;j<prime.size() && i*prime[j] < 100005 && prime[j] <= lpf[i] ;j++){
                lpf[i * prime[j]] = prime[j];
                // i*prime[j] < 100005, else out of bounds
                // lpf[i * prime[j]] = prime[j] means lpf[i] >= prime[j]
                // so, just remember lpf[i * prime[j]] = prime[j], and write the 2 conditions 
                // isPrime[i*prime[j]] = 0;
            }
        }
    }

    bool canTraverseAllPairs(vector<int>& a) {
        // bfs/union find both are possible

        // just gcd > 1 => prime factor matching is enough lol
        // took me 15 mins to get this idea, not good

        // my method : make edges if a[i] , a[j] have a common "prime" factor
        // but storing it in a map(the idx which have a common prime factor) makes it slower
        // checking: if all i=0 to n-1 are visited or not

        // better method : make edges from a[i] to all its prime factors
        // then check if all a[i]'s are visited 
        

        int n = a.size();
        if(n == 1){
            return true;
        }

        memset(vis , 0 , sizeof(vis));
        memset(lpf , 0 , sizeof(lpf));
        prime.clear();

        lin();

        for(int i=0;i<n;i++){
            adj[i].clear();
        }

        // map<int , vector<int>> m; // prime , {indices which it divides}
        
        for(int i=0;i<n;i++){
            if(a[i] == 1){
                return false;
            }
            int num = a[i];
            while(1){
                int pr = lpf[num];
                // m[pr].push_back(i);
                adj[a[i]].push_back(pr);
                adj[pr].push_back(a[i]);
                while(num % pr == 0){
                    num /= pr;
                }
                if(num == 1)break;
            }
        }

        // for(auto it : m){
        //     vector<int> idx = it.second;
        //     // no need to make all nC2 of them, just need to check connectivity
        //     for(int i=0;i<idx.size()-1;i++){
        //         adj[idx[i]].push_back(idx[i+1]);
        //         adj[idx[i+1]].push_back(idx[i]);
        //     }
        // }

        dfs(a[0] , -1);
        for(int i=0;i<n;i++){
            if(vis[a[i]] == 0)return false;
        }
        return true;
    }
};
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