<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 disjointSet{
private:
    int n;
    vector<int>rank;
    vector<int>parent;
public:
    disjointSet(int n){
        this -> n = n;
        parent.resize(n , 0);
        rank.resize(n , 1);
        for(int i = 0 ; i < n ; i++)
            parent[i] = i;
    }
    int find(int a){
        if(parent[a] == a)
            return a;
        return parent[a] = find(parent[a]);
    }
    void Union(int a , int b){
        int a_rep = find(a);
        int b_rep = find(b);

        if(a_rep == b_rep)return ;

        if(rank[a_rep] < rank[b_rep]){
            parent[a_rep] = b_rep;
            rank[b_rep] += rank[a_rep];
        }else{
            parent[b_rep] = a_rep;
            rank[a_rep] += rank[b_rep];
        }
    }
};

class Solution {
public:
    void findFactors(int num , int i , unordered_map<int,vector<int>>& mp){
        for(int j  = 2 ; j* j <=num ; j++){
            if(num % j == 0){
                mp[j].push_back(i);
                while(num % j == 0){
                    num /= j;
                }
            }
        }
        if(num > 1){
            mp[num].push_back(i);
        }
    }
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        disjointSet ds(n);
        unordered_map<int , vector<int>>mp;
        for(int i = 0 ; i < n ; i++){
            findFactors(nums[i] , i , mp);
        }   
        for(auto pr : mp){
            vector<int>indexes = pr.second;
             cout << pr.first << " : ";
            for(int i = 1 ; i < indexes.size() ; i++){
                ds.Union(indexes[i] , indexes[i-1]);
                cout<<indexes[i-1] << " " ;
            }
            cout<<indexes[indexes.size()-1] << endl;
        }
        set<int>components;
        for(int i = 0 ; i < n ; i++){
            cout << ds.find(i) << " ";
            components.insert(ds.find(i));
        }
        return components.size() == 1;
    }
};
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