<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 for union find
class DS
{
    public:
  vector<int> parent, size;

  DS(int n)
  {
    parent.resize(n + 1);
    size.resize(n + 1);
    for (int i = 0; i <= n; i++)
    {
      parent[i] = i;
      size[i] = 1;
    }
  }

  int find(int node)
  {
    if (node == parent[node])
      return node;
    return parent[node] = find(parent[node]);
  }

  void union_size(int u, int v)
  {
    int ulp_u = find(u);
    int ulp_v = find(v);
    if (ulp_u == ulp_v)
      return;
    if (size[ulp_u] < size[ulp_v])
    {
      parent[ulp_u] = ulp_v;
      size[ulp_v] += size[ulp_u];
    }
    else
    {
      parent[ulp_v] = ulp_u;
      size[ulp_u] += size[ulp_v];
    }
  }
};

class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        if(nums.size()==1){
            return true;
        }

        //adding each value to sets of its factors
        unordered_map<int,vector<int>> mp;
        for(int i=0 ; i<nums.size() ; i++){
            if(nums[i]==1){
                return false;
            }
            for(int j=2 ; j*j<=nums[i] ; j++){
                if(nums[i]%j==0){
                    mp[j].push_back(i);
                    int k=nums[i]/j;
                    if(k==j){
                        continue;
                    }
                    mp[k].push_back(i);
                }
            }
            mp[nums[i]].push_back(i);
        }
        
        //combining/ union of values having same factor
        DS ds(nums.size());
        for(auto m : mp ){
            int x=m.first;;
            for(int i=0 ; i<mp[x].size()-1 ; i++){
                ds.union_size(ds.find(mp[x][i]),ds.find(mp[x][i+1]));
            }
        }
        int count=0;
        for(int i=0 ; i<nums.size() ; i++){
            if((ds.find(i)==i)){
                count++;
            }
            if(count>1){
              return false;
            }
        }

        //if all belong to same group
        if(count==1){
            return true;
        }

        return false;
    }
};
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