<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 DSU {
    public:
    unordered_map<int,int>parent , rank;

    DSU(){
    }

    void MakeSet(int x){
        parent[x] = x;
        rank[x] = 0;
    }

    int Find(int x){
        return parent[x] = parent[x] ==  x ? x : Find(parent[x]);
    }

    bool Union(int x , int y){

        int xp = Find(x);
        int yp = Find(y);

        if(xp != yp){
        rank[xp] < rank[yp] ? parent[xp] = yp : parent[yp] = xp;
        rank[xp] += rank[xp] == rank[yp];
        }

        return xp != yp;
    }
};
class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {  

        int n = nums.size();

        if(n == 1){
            return true;
        }

        if(find(nums.begin() , nums.end() , 1) != nums.end()){
            return false;
        }

        DSU dsu = DSU();

        for(int i = 0 ; i < n ; ++i){ 
            int elem = nums[i];
            // vector<int>primes;
            int parent = 0;
            int first = 1;
            for(int j = 2 ; j*j <= nums[i] ; ++j){
                if(elem%j == 0){
                    if(dsu.parent[j] == 0){
                    dsu.MakeSet(j);
                    }
                    if(first){
                        parent = j;
                        first = 0;
                    }
                    dsu.Union(parent , j);
                    // primes.push_back(j);
                    while(elem%j == 0){
                        elem /= j;
                    }
                }
            }
            if(elem != 1){
            if(dsu.parent[elem] == 0){
            dsu.MakeSet(elem);
            }
            if(first){
                parent = elem;
                first = 0;
            }
            dsu.Union(parent , elem);
            // primes.push_back(elem);
            }

            // for(int&j : primes){
            //     dsu.Union(primes[0] , j);
            // }
        }

        int root = dsu.Find(dsu.parent.begin()->second);

        for(auto&[el , par] : dsu.parent){
            if(dsu.Find(par) != root){
                return false;
            }
        }

        return true;

    }
};
86737