<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> getPrime(int n) 
    {   
        vector<int> prime;
        if(n % 2 == 0){
            prime.push_back(2);
            while (n % 2 == 0) { 
                n = n/2; 
            } 
        }
        // n must be odd at this point. So we can skip 
        // one element (Note i = i +2) 
        for (int i = 3; i <= sqrt(n); i = i + 2) { 
            if(n % i == 0){
                prime.push_back(i);
                while (n % i == 0) { 
                    n = n/i;
                } 
            }
        } 
 
        // This condition is to handle the case when n 
        // is a prime number greater than 2 
        if (n > 2) 
            prime.push_back(n);
        return prime;
    } 
    void dfs(vector<int> graph[],int node,vector<int>&vis){
        vis[node]=1;
        for(auto child : graph[node]){
            if(vis[child]==0){
                dfs(graph,child,vis);
            }
        }
    }
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        if(n==1)
            return true;
        vector<int> graph[n];
        unordered_map<int,int> map;
        for(int i=0;i<n;i++){
            if(nums[i]==1)
                return false;
            vector<int>  prime = getPrime(nums[i]);
            for(auto x : prime){
                if(map.find(x)!=map.end()){
                    int u = map[x];
                    int v = i;
                    graph[u].push_back(v);
                    graph[v].push_back(u);
                }
                else map[x]=i;
            }
        }
        vector<int> vis(n,0);
        int count = 0;
        for(int i=0;i<n;i++){
            if(vis[i]==0){
                count++;
                dfs(graph,i,vis);
            }
        }
        return count == 1;
    }
};
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