<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 {
    
    int par[100007];int size[100007];
    void dsu(int n){
        for(int i=1; i<=n; i++) par[i]=i,size[i]=1;
    }

    int find(int x){
        if( x==par[x]) return x;
        return par[x]=find(par[x]);
    }

    void unite(int x, int y){
        int a=find(x),b=find(y);
        if( a != b){
            if( size[a]<size[b]) swap(a,b);
            par[b]=a;
            size[a] +=size[b];
        }
    }
public:
    bool canTraverseAllPairs(vector<int>& v1) {
        if( v1.size()==1) return true;
        unordered_map<int,int> mp,vis;
        for(int i=0; i<v1.size(); i++) mp[v1[i]]=1;
        if( mp.find(1) != mp.end() ) return false;
        int n=*max_element(v1.begin() , v1.end());
        dsu(n);
        for(int i=2; i<=n; i++){
            if( vis[i]==1) continue;
            for(int j=2*i ; j<=n; j=j+i){
                if( mp[j]==1){
                    unite(i,j);
                }
                vis[j]=1;
            }
        }
        
        unordered_map<int,int> temp; int cnt=0;
        for(int val : v1){
            int a=find(val); 
            if( temp[a]==0) cnt++;
            temp[a]=1;
        }
        if( cnt ==1) return true;
        return  false;
    }
};
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