<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:
    vector<int>parent,rank,size;
    void dsu(int n){
        rank.resize(n+1,0);
        parent.resize(n+1);
        size.resize(n+1,1);
        for(int i=0;i<=n;i++){
            parent[i]=i;
        }
    }
    int findpar(int node){
        if(node==parent[node])return node;
        return parent[node]=findpar(parent[node]);
    }

    void unionsize(int u,int v){
        int pu=findpar(u);
        int pv=findpar(v);
        if(pu==pv)return;
        else  if(size[pu]<size[pv]){
        parent[pu]=pv;
        size[pv]+=size[pu];
        }
        else{
            parent[pv]=pu;
            size[pu]+=size[pv];
        }
}
    bool canTraverseAllPairs(vector<int>& nums) {
       int n = nums.size();
        dsu(n);
        map<int, vector<int>> m;

        for (int i = 0; i < n; i++) {
            if (nums[i] == 1) continue;
            for (int j = 2; j * j <= nums[i]; j++) {
                if (nums[i] % j == 0) {
                    m[j].push_back(i);
                    if (j != nums[i] / j) {
                        m[nums[i] / j].push_back(i);
                    }
                }
            }
            if (nums[i] > 1) m[nums[i]].push_back(i);
        }

        for (auto it : m) {
            vector<int>v = it.second;
            if (!v.empty()) {
                int el = v[0];
                for (int i = 1; i < v.size(); i++) {
                    unionsize(el, v[i]);
                }
            }
        }

        int par = findpar(0);
        for (int i = 1; i < n; i++) {
            if (findpar(i) != par) return false;
        }
        return true;
    }
};
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