<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:
    using bitset = std::vector<std::int8_t>;

    std::vector<int> generatePossibleDivizors(int n) {
        n /= 2;
        auto visit = bitset(n);
        
        for (auto j = 4; j < n; j += 2) {
            visit[j] = true;
        }
        for (auto i = 3; i < n / 3 + 1; i += 2) {
            if (visit[i]) {
                continue;
            }
            for (auto j = 3 * i; j < n; j += 2 * i) {
                visit[j] = true;
            }
        }
        
        auto ret = std::vector<int>{};
        for (auto i = 2; i < n; ++i) {
            if (not visit[i]) {
                ret.push_back(i);
            }
        }
        return ret;
    }

    int find_parent(auto& container, int what) {
        if (container[what] == 0) {
            container[what] = what + 1;
        } else if (container[what] != what + 1) {
            container[what] = find_parent(container, container[what] - 1);
        }
        return container[what];
    }

    bool canTraverseAllPairs(const std::vector<int>& nums) {
        const auto size = nums.size();
        if (size < 2) {
            return true;
        }
        auto divisors = generatePossibleDivizors(100000); // size 5133
        auto cluster_parents = std::vector<int>(divisors.size(), 0);
        auto big_num = 0;
        auto big_num_counts = 0;

        for (auto x : nums) {
            auto ok = false;
            auto current_cluster = std::set<int>();

            for (auto d = 0; d < divisors.size(); ++d) {
                const auto div = divisors[d];
                if (x % div == 0) {
                    ok = true;
                    current_cluster.insert(find_parent(cluster_parents, d));
                    do {
                        x /= div;
                    } while(x % div == 0);
                }
                if (x < div) {
                    break;
                }
            }
            if (not ok) {
                if (x == 1) {
                    return false; 
                } else if (big_num == 0) {
                    big_num = x;
                    ++big_num_counts;
                } else if (big_num != x) {
                    return false;
                } else {
                    ++big_num_counts;
                }
            } else {
                auto new_parent = *current_cluster.begin(); // guaranteed to exist
                std::for_each(std::next(current_cluster.begin()), current_cluster.end(),
                [new_parent, &cluster_parents](auto former_parent) {
                    cluster_parents[former_parent - 1] = new_parent;
                });
            }
        }
        if (big_num != 0) {
            return big_num_counts == size;
        }
        
        auto first_non_zero = 0;
        for (auto p : cluster_parents) {
            if (p) {
                if (not first_non_zero) {
                    first_non_zero = find_parent(cluster_parents, p - 1);
                } else if (first_non_zero != p) {
                    if (find_parent(cluster_parents, p - 1) != first_non_zero) {
                        return false;
                    }

                }
            }
        }
        return true;
    }
};
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