<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 UnionFind {
public:
    UnionFind(int size) : parent(size), rank(size, 1) {
        for (int i = 0; i < size; ++i) {
            parent[i] = i;
        }
    }

    int find(int u) {
        if (parent[u] != u) {
            parent[u] = find(parent[u]);
        }
        return parent[u];
    }

    void unionSets(int u, int v) {
        int rootU = find(u);
        int rootV = find(v);
        if (rootU != rootV) {
            if (rank[rootU] > rank[rootV]) {
                parent[rootV] = rootU;
            } else if (rank[rootU] < rank[rootV]) {
                parent[rootU] = rootV;
            } else {
                parent[rootV] = rootU;
                ++rank[rootU];
            }
        }
    }

private:
    vector<int> parent;
    vector<int> rank;
};

vector<int> getPrimeFactors(int n) {
    vector<int> factors;
    if (n % 2 == 0) {
        factors.push_back(2);
        while (n % 2 == 0) {
            n /= 2;
        }
    }
    for (int i = 3; i <= sqrt(n); i += 2) {
        if (n % i == 0) {
            factors.push_back(i);
            while (n % i == 0) {
                n /= i;
            }
        }
    }
    if (n > 2) {
        factors.push_back(n);
    }
    return factors;
}

class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        UnionFind uf(n);

        unordered_map<int, vector<int>> primeToIndex;

        for (int i = 0; i < n; ++i) {
            vector<int> primeFactors = getPrimeFactors(nums[i]);
            for (int prime : primeFactors) {
                primeToIndex[prime].push_back(i);
            }
        }

        for (auto& [prime, indices] : primeToIndex) {
            for (size_t i = 1; i < indices.size(); ++i) {
                uf.unionSets(indices[0], indices[i]);
            }
        }

        int root = uf.find(0);
        for (int i = 0; i < n; ++i) {
            if (uf.find(i) != root) {
                return false;
            }
        }

        return true;
    }
};
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