<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:
    bool canTraverseAllPairs(vector<int>& nums) {
        unordered_set<int> nums_set{};

        unordered_map<int, vector<int>> prime_to_idxs{};
        vector<vector<int>> idx_to_primes(ssize(nums));

        vector<bool> visited(ssize(nums), false);
        
        auto update_graph = [&](int idx) -> void {
            int num = nums[idx];
            auto [_, inserted] = nums_set.insert(num);
            if (!inserted) {
                visited[idx] = true;
                return;
            }

            vector<int> &primes_list = idx_to_primes[idx];

            for (long long i = 2; i * i <= num; i++) {
                bool found_prime = false;
                while (num % i == 0) {
                    found_prime = true;
                    num /= i;
                }

                if (found_prime) {
                    primes_list.push_back(i);
                    prime_to_idxs[i].push_back(idx);
                }
            }

            if (num != 1) {
                primes_list.push_back(num);
                prime_to_idxs[num].push_back(idx);
            }
        };

        for (int i = 0; i < ssize(nums); i++) {
            if (nums[i] == 1 && ssize(nums) > 1) return false;
            update_graph(i);
        }

        stack<int> st{};
        visited[0] = true;
        st.push(0);

        while (!st.empty()) {
            int curr = st.top(); st.pop();

            for (int prime : idx_to_primes[curr]) {
                for (int neigh : prime_to_idxs[prime]) {
                    if (!visited[neigh]) {
                        visited[neigh] = true;
                        st.push(neigh);
                    }
                }
            }
        }

        return find(begin(visited), end(visited), false) == end(visited);
    }
};
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