<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 Disjoint{
    vector<int>par,size,rank;
public:
    Disjoint(int n)
    {
        par.resize(n+1);
        size.resize(n+1,1);
        rank.resize(n+1,0);
        for(int i=0 ; i<=n ; i++)par[i]=i;
    }
    int findPar(int a)  
    {
        if(a==par[a])return a;
        return par[a]=findPar(par[a]);
    }
    void unionBySize(int a,int b)
    {
        int parA=findPar(a);
        int parB=findPar(b);
        if(parA==parB)return;
        if(size[parA]<size[parB])
        {
            par[parA]=parB;
            size[parB]+=size[parA];
        }
        else
        {
            par[parB]=parA;
            size[parA]+=size[parB];   
        }
    }
};
class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n=nums.size();
        if(n==1)return true;
        int maxi=*max_element(begin(nums),end(nums));
        vector<int>f(maxi+1,0);
        for(int &x : nums)f[x]++;

        if(f[1])return false;
        Disjoint dsu(maxi+1);

        int p=-1;
        for(int i=2 ; i<=maxi ; i++){
            for(int j=i ; j<=maxi ; j+=i){
                if(f[j]==0)continue;
                dsu.unionBySize(i,j);
                p=dsu.findPar(j);
            }
        }
        for(int i=2 ; i<=maxi ; i++){
            if(f[i] && dsu.findPar(i)!=p)return false;
        }
        return true;
    }
};
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