<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>

const int N=1e5+7;
vector<int>parent(N,0);
vector<int>sz(N,0);
class Solution {
public:
    int find(int v){
       if(v==parent[v])return v;
       return parent[v]= find(parent[v]);
    }
    void union_2gr(int a,int b){
       a=find(a);
       b=find(b);
    
       if(a!=b){
        //union by size
        if(sz[a]<sz[b])swap(a,b);
         parent[b]=a;
         sz[a]+=sz[b];
       }  
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        map<int,vector<int>>m1;
        int n=nums.size();
        for(int i=0;i<n;i++)parent[i]=i;

        for(int i=0;i<n;i++){
            int k=nums[i];
            for(int j=2;j*j<=k;j++){
                if(k%j==0){
                    m1[j].push_back(i);
                }
                while(k%j==0 && k>0)k/=j;
            }
            if(k>=2)m1[k].push_back(i);
        }
        for(auto x:m1){
            auto v=x.second;
           for(int i=1;i<v.size();i++){
               union_2gr(v[0],v[i]);
           }
        }
        int ct=0;
        for(int i=0;i<n;i++){
            if(parent[i]==i)ct++;
        }
        return ct==1;
    }
};
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