<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 {
public:

    struct DisjointSets {

        int n;
        vector<int> parent;
        vector<bool> touched;
        vector<int> rank;

        DisjointSets(int n) {
            this->n = n;
            parent.assign(n, 0);
            iota(parent.begin(), parent.end(), 0);
            rank.assign(n, 1);
            touched.assign(n, false);
        }

        void touch(int i) {
            touched[i] = true;
        }

        int find(int i) {

            if (i == parent[i]) return i;

            int res = find(parent[i]);
            parent[i] = res;

            return res;

        }

        void merge(int i, int j) {
            
            int I = find(i);
            int J = find(j);
            if (I == J) return;

            if (rank[I] > rank[J]) {
                parent[J] = I; 
            } else {
                parent[I] = J; 
            }

        }

        bool connected() {
            int onlyParent = -1;
            for (int i = 0; i < parent.size(); i++) {
                if (touched[i]) {
                    if (onlyParent == -1) onlyParent = find(i);
                    if (find(i) != onlyParent) return false;
                }
            }
            return true;
        }

    };

    vector<int> getPrimes(int n) {

        vector<int> sieve(n+1, 0);
        iota(sieve.begin(), sieve.end(), 0);

        vector<int> res;

        for (int i = 2; i <= n; i++) {
            
            if (sieve[i] == -1) continue;

            res.push_back(i);
            for (int j = i; j <= n; j+=i) {
                sieve[j] = -1;
            }

        }

        return res;

    }

    bool canTraverseAllPairs(vector<int>& nums) {

        set<int> s(nums.begin(), nums.end());
        if (s.find(1) != s.end()) {
            return nums.size() == 1;
        }
        nums.assign(s.begin(), s.end());

        vector<int> primes = getPrimes(*prev(s.end()));

        DisjointSets ds(primes.size());

        for (int num : nums) {
            
            const int limit = num/2+1;

            vector<int> factorIndices;
               
            for (int i = 0; i < primes.size() and primes[i] < limit; i++) {
                if (num % primes[i] == 0) {
                    ds.touch(i);
                    factorIndices.push_back(i);
                }
            }

            const int numIndexInPrimes = lower_bound(primes.begin(), primes.end(), num) - primes.begin();
            if (numIndexInPrimes < primes.size() and primes[numIndexInPrimes] == num) {
                ds.touch(numIndexInPrimes);
                factorIndices.push_back(numIndexInPrimes);
            }

            // cout << "------" <<endl;
            // cout << num << endl;
            // for (int ind : factorIndices) cout << ind << endl;

            for (int i = 0; i < factorIndices.size(); i++) {
                for (int j = i+1; j < factorIndices.size(); j++) {
                    ds.merge(factorIndices[i], factorIndices[j]);
                }
            }

        }



        return ds.connected();
        
    }
};
113962