<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>

struct UnionFind {
    vector<int> parent, rank;
    UnionFind(int n) {
        parent.resize(n); rank.resize(n, 0);
        iota(parent.begin(), parent.end(), 0);
    }
    int find(int x) { return x == parent[x] ? x : parent[x] = find(parent[x]); }
    bool isConnected(int x, int y) { return find(x) == find(y); }
    void unite(int x, int y) {
        if (isConnected(x, y)) return;
        int px = find(x), py = find(y);
        if (rank[px] == rank[py]) {
            ++rank[px];
            parent[py] = px;
        } else if (rank[px] < rank[py]) {
            parent[px] = py;
        } else {
            parent[py] = px;
        }
    }
};
class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        UnionFind uf(n);
        unordered_map<int, int> um;
        int max_num = *max_element(nums.begin(), nums.end());
        vector<int> sieve(max_num+1);
        iota(sieve.begin(), sieve.end(), 0);
        for (int i = 2; i <= max_num; ++i) {
            if (sieve[i] == i) {
                for (int j = i; j <= max_num; j += i) {
                    sieve[j] = i;
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            int curr = nums[i];
            while (curr > 1) {
                int p = sieve[curr];
                if (um.contains(p)) uf.unite(i, um[p]);
                else um[p] = i;
                while (curr%p == 0) curr /= p;
            }
        }
        for (int i = 0; i < n; ++i) {
            if (!uf.isConnected(0, i)) return false;
        }
        return true;
    }
};
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