<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>


struct dsu {
    vector<int> par, rk;
    int connected;

    void init(int n) {
        par.resize(n);
        rk.resize(n, 1);
        iota(par.begin(),par.end(), 0);
        connected = n;
    }

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

    void unite(int a, int b) {
        a = getp(a);
        b = getp(b);
        if (a == b) return;
        connected--;
        if (rk[a] > rk[b]) swap(a, b);
        rk[b] += rk[a];
        par[a] = b;
    }
};


class Solution {
public:
    bool canTraverseAllPairs(vector<int>& a) {
        if(a.size()==1) return true;
        dsu ds;
        ds.init(1e5+1);
        for(auto x:a){
            if(x==1) return false;
            int y=x;
            vector<int> p;
            for(int i=2;i*i<=y;i++){
                if(y%i) continue;
                p.push_back(i);
                while(y%i==0) y/=i;
            }
            if(y!=1) p.push_back(y);
            for(int i=1;i<p.size();i++) ds.unite(p[0],p[i]);
        }
        int f = -1;
        for(auto x:a){
            int y = x;
            for(int i=2;i*i<=y;i++){
                if(y%i) continue;
                // cout<<x<<" "<<i<<" "<<ds.getp(i)<<endl;
                if(f==-1) f = ds.getp(i);
                else if(ds.getp(i)!=f) return false;
                while(y%i==0) y/=i;
            }
            if(y!=1) {
                // cout<<y<<" "<<ds.getp(y)<<endl;
                if(f==-1) f = ds.getp(y);
                else if(ds.getp(y)!=f) return false;
            }
        }
        return true;
    }
};
319662