<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 {
    vector<int> parent, rank;

public:
    DisjointSet(int n) {
        parent.resize(n + 1);
        rank.resize(n + 1, 0);

        for(int i = 0; i <= n; i++) {
            parent[i] = i;
        }
    }

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

    void unionBySize(int u, int v) {
        int root_u = findUpar(u);
        int root_v = findUpar(v);

        if (root_u != root_v) {
            if (root_u < root_v) {
                parent[root_v] = root_u;
            } else {
                parent[root_u] = root_v;
            }
        }
    }
};

class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        if(n == 1) return true;
        sort(nums.begin(), nums.end());
        int mx = nums[n-1];
        DisjointSet ds(mx + 1);

        for(int i = 0; i < n; i++) {
            if(nums[i] == 1) return false;
            for(int j = 2; j * j <= nums[i]; j++) {
                if(nums[i] % j == 0) {
                    ds.unionBySize(nums[i], j);
                    ds.unionBySize(nums[i], nums[i] / j);
                }
            }
        }

        for(int i = 0; i < n-1; i++){ 
            if(ds.findUpar(nums[i]) != ds.findUpar(nums[i+1])) return false;
        }

        return true;
    }
};
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