<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 {
private:
    int findParent(vector<int> &parent, int u){
        return parent[u]=(parent[u]==u)?u:findParent(parent, parent[u]);
    }
    void join(vector<int> &parent, vector<int> &size, int u, int v){
        int pu=findParent(parent, u), pv=findParent(parent, v);
        if(pu==pv) return;
        if(size[pu]<size[pv]){
            size[pv]+=size[pu];
            parent[pu]=pv;
        }
        else{
            size[pu]+=size[pv];
            parent[pv]=pu;
        }
    }
public:
    bool canTraverseAllPairs(vector<int>& nums){
        int maxi=INT_MIN, n=nums.size(), cnt=0;
        if(n==1) return true;
        unordered_map<int, int> mp;
        for(int i=0;i<n;i++){
            if(mp.find(nums[i])!=mp.end()) continue;
            if(nums[i]==1) return false;
            maxi=max(maxi, nums[i]);
            mp[nums[i]]=i;
            cnt++;
        }
        // getting cnt unique numbers. last me agar aesi koi node hai jo cnt nodes ki parent hai then we are good;
        // mapped all indexes to numbers, could have done directly also but this is better;
        vector<bool> isPrime(maxi+1, true);
        vector<int> parent(n), size(n, 1);
        iota(parent.begin(), parent.end(), 0);
        // applying sieve me hi dsu ka logic saath me;
        for(int i=2;i<=maxi;i++){
            if(!isPrime[i]) continue;
            int merger=-1;
            for(int j=i;j<=maxi;j+=i){
                isPrime[j]=false; // galti se 2, 3, 5... bhi not prime marked but who cares;
                if(mp.find(j)==mp.end()) continue; // element not found toh no need to join;
                if(merger==-1) merger=mp[j]; // find ek ele jo hai nums me and use it to merge all of them; 
                join(parent, size, merger, mp[j]);
            }
        }
        unordered_map<int, int> mp2;
        for(int i=0;i<n;i++){
            if(++mp2[findParent(parent, i)]==cnt) return true; // koi bhi node cnt nodes ki parent hai toh we found the answer;
        }
        return false; // else false;
    }
};
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