<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:
    struct dsu
    {
        vector<int> sz;
        vector<int> p;
        void initialize(int n)
        {
            sz.resize(n,1);
            p.resize(n,0);
            for(int i=0; i<n; i++) p[i]=i;
        }
        int get(int u)
        {
            if(p[u]!=u) p[u]=get(p[u]);
            return p[u];
        }
        void join(int x, int y)
        {
            x=get(x);
            y=get(y);
            if(x==y) return;
            if(sz[y]>sz[x]) swap(x,y);
            sz[x]+=sz[y];
            p[y]=x;
            return;
        }

    };

    bool canTraverseAllPairs(vector<int>& nums1) {
        int n=nums1.size();
        set<int> s2;
        if(n==1) return true;
        for(int i=0; i<n; i++) if(nums1[i]==1) return false;
        for(int i=0; i<n; i++) s2.insert(nums1[i]);
        vector<int> nums(s2.begin(),s2.end());
        n=s2.size();
        vector<int> v(100005,-1);
        for(int i=0; i<nums.size(); i++) v[nums[i]]=i;
        vector<int> p(100005,0);
        dsu d;
        d.initialize(n);
        int k=100005;
        for(int i=2; i<k; i++)
        {
            if(p[i]==1) continue;
            int l=-1;
            for(int j=i; j<k; j+=i)
            {
                p[j]=1;
                if(v[j]==-1) continue;
                if(l!=-1)
                {
                    // cout<<l<<" "<<v[j]<<endl;
                    d.join(l,v[j]);
                }
                l=v[j];


            }
        }
        set<int> st;
        for(int i=0; i<n; i++) st.insert(d.get(i));
        // cout<<st.size();
        if(st.size()==1) return true;
        return false;
    }
};
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