<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:
  int findpar(int node,unordered_map<int,int>&parent) {
        if (node == parent[node])
            return node;
        return parent[node] = findpar(parent[node],parent);
    }
     void uni(int u, int v,unordered_map<int,int>&parent,unordered_map<int,int>&rank) {
        int ulp_u = findpar(u,parent);
        int ulp_v = findpar(v,parent);
        if (ulp_u == ulp_v) return;
        if (rank[ulp_u] < rank[ulp_v]) {
            parent[ulp_u] = ulp_v;
        }
        else if (rank[ulp_v] < rank[ulp_u]) {
            parent[ulp_v] = ulp_u;
        }
        else {
            parent[ulp_v] = ulp_u;
            rank[ulp_u]++;
    }
    }
    bool canTraverseAllPairs(vector<int>& nums) {
        unordered_map<int,int>parent;
        unordered_map<int,int>rank;
        int n=nums.size();
        if(n==1)return true;
         for (int i = 0; i < n; i++) {
            if(nums[i]==1)return false;
            parent[nums[i]] = nums[i];
            rank[nums[i]] = 1;
        }
        for(int i=0;i<n;i++){
            if(nums[i]==1)continue;
            int sq=sqrt(nums[i]);
            for(int j=2;j<=sq;j++){
                if(nums[i]%j==0){
                 if(parent.find(j)==parent.end())parent[j]=j;
                 int a =findpar(j,parent);
                 int b =findpar(nums[i],parent);
                 if(a!=b)uni(a,b,parent,rank);
                  int c=nums[i]/j;
                 if(parent.find(c)==parent.end())parent[c]=c;
                 if(gcd(c,nums[i])>1){
                   int a =findpar(c,parent);
                   int b =findpar(nums[i],parent);
                   if(a!=b)uni(a,b,parent,rank);
                 }
            }
            }
        }
        int p=-1;
        int c=0;
        for(int i=0;i<n;i++){
            int pa=findpar(nums[i],parent);
            if(pa!=p){
                c++;
                p=pa;
            }
        }
        return c==1;
    }
};
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