<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:
    void make(int node, vector<int>& parent, vector<int>& s){
        s[node]=1;
        parent[node]=node;
    }
    int find(int node,vector<int>& parent){
        if(node!=parent[node]){
            return parent[node]=find(parent[node],parent);
        }
        return node;
    }
    void union_set(int a, int b, vector<int>& parent, vector<int>& s){
        if(parent[a]==parent[b]){
            return;
        }
        a=parent[a],b=parent[b];
        if(s[a]>=s[b]){
            s[a]+=s[b];
            parent[b]=a;
        }
        else{
            s[b]+=s[a];
            parent[a]=b;
        }
    }
    bool canTraverseAllPairs(vector<int>& nums) {
        int n=nums.size();
        int maxi=0;
        map<int,int> mp;
        for(int i=0;i<n;i++){
            mp[nums[i]]++;
            maxi=max(maxi,nums[i]);
        }
        if(mp.find(1)!=mp.end()){
            if(nums.size()==1){
                return true;
            }
            return false;
        }
        vector<int> s(maxi+1,0),parent(maxi+1,-1);
        for(int i=1;i<=maxi;i++){
            make(i,parent,s);
        }
        map<int,int> fac;
        for(int i=2;i<=maxi;i++){
            for(int j=i;j<=maxi;j+=i){
                if(mp.find(j)!=mp.end()){
                    fac[i]++;
                }
            }
        }
        for(int j=2;j<=maxi;j++){
            for(int i=2*j;i<=maxi;i+=j){
                if(fac.find(j)!=fac.end() && fac.find(i)!=fac.end()){
                    union_set(i,j,parent,s);
                }
            }
        }
        map<int,int> check;
        for(auto it:mp){
            check[find(it.first,parent)]++;
        }
        return check.size()==1;
    }
};
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