<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 UF {
    private:
        std::vector<int> parent;

    public:
        UF(int len) : parent(len, -1) {}

        int find(int a) {
            if (parent[a] >= 0) {
                return parent[a] = find(parent[a]);
            }
            return a;
        }

        bool unionSet(int a, int b) {
            int pa = find(a);
            int pb = find(b);

            if (pa == pb) return false;

            if (pa < pb) { // Merge by rank heuristic could be applied here
                parent[pa] += parent[pb]; // Merge sizes (or ranks)
                parent[pb] = pa;
            } else {
                parent[pb] += parent[pa]; // Merge sizes (or ranks)
                parent[pa] = pb;
            }

            return true;
        }
};


class Solution {
public:
    vector<int> getPrimeFactors(int n) {
        vector<int> factors;
        // Check for number of 2s
        while (n % 2 == 0) {
            factors.push_back(2);
            n /= 2;
        }
        // Check for odd factors
        for (int i = 3; i <= sqrt(n); i += 2) {
            while (n % i == 0) {
                factors.push_back(i);
                n /= i;
            }
        }
        if (n > 2) factors.push_back(n);  // n is prime number greater than 2
        return factors;
    }
    
    bool canTraverseAllPairs(vector<int>& A) {
        int N = A.size();

        map<int, vector<int>> memo;
        map<int, vector<int>> primeToIdx;
        UF uf(N);

        for(int i=0; i<N; i++) {
            if(!memo.count(A[i])) {
                memo[A[i]] = getPrimeFactors(A[i]);
            }
            auto primes = memo[A[i]];
            for(int p : primes) {
                if(!primeToIdx[p].empty()) {
                    uf.unionSet(primeToIdx[p].front(), i);
                }
                primeToIdx[p].push_back(i);
            }
        }

        for(int i=1; i<N; i++) {
            if(uf.find(0) != uf.find(i)) return false;
        }

        return true;
    }
};
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