<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 {
    class UnionFind {
        vector<int> reps, szs;
    public:
        UnionFind(int n) : reps(n), szs(n, 1) {
            iota(begin(reps), end(reps), 0);
        }

        int find(int i) {
            if (reps[i] != i) reps[i] = find(reps[i]);
            return reps[i];
        }

        int size(int i) {
            if (reps[i] == i) return szs[i];
            return size(find(i));
        }

        void merge(int i1, int i2) {
            int rep1 = find(i1), rep2 = find(i2);
            if (rep1 == rep2) return;

            if (szs[rep1] > szs[rep2]) {
                reps[rep2] = rep1;
                szs[rep1] += szs[rep2];
            }
            else {
                reps[rep1] = rep2;
                szs[rep2] += szs[rep1];
            }
        }
    };
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        unordered_map<int, vector<int>> numsWithFactor;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int num = nums[i];
            int factor = 2;
            while (factor * factor <= num) {
                if (num % factor == 0) {
                    numsWithFactor[factor].push_back(i);
                    while (num % factor == 0) num /= factor;
                }
                ++factor;
            }
            if (num > 1) numsWithFactor[num].push_back(i);
        }

        UnionFind uf(n);
        for (auto& [_, indices] : numsWithFactor) {
            int m = indices.size();
            for (int i = 0; i < m-1; ++i) {
                uf.merge(indices[i], indices[i+1]);
            }
        }
        return uf.size(0) == n;
    }
};
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