<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 {
    int parent[110001];
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        int max_num = *max_element(nums.begin(), nums.end());
        vector<int> primes = get_primes(max_num);
        memset(parent, -1, sizeof(parent));

        unordered_map<int, int> prime_to_index;
        int size = primes.size();
        for(int i=0; i<size; i++) prime_to_index[primes[i]] = i;

        for(int i=0; i<n; i++){
            int num = nums[i];
            for(int p_id=0; p_id<size; p_id++){
                int prime = primes[p_id];
                if(prime>num) break;
                if(prime*prime>num){
                    __union(prime_to_index[num], i+size);
                    break;
                }
                if(num%prime!=0) continue;

                while(num%prime==0) num/=prime;

                __union(p_id, i+size);
            }
        }

        for(int i=size+1; i<n+size; i++){
            int x = __find(size);
            int y = __find(i);
            if(__find(size)!=__find(i)) return false;
        }
        return true;
    }

    int __find(int x){
        if(parent[x]==-1) return x;
        return parent[x] = __find(parent[x]);
    }

    void __union(const int x, const int y){
        int px = __find(x);
        int py = __find(y);
        if(px==py) return;

        if(px<py) parent[py] = px;
        else parent[px] = py;
    }


    vector<int> get_primes(int n){
        bool is_prime[n+1];
        memset(is_prime, true, sizeof(is_prime));

        for(int i=2; i*i<=n; i++){
            if(!is_prime[i]) continue;

            for(int t=i*2; t<=n; t+=i) is_prime[t] = false;
        }

        vector<int> primes;
        for(int i=2; i<=n; i++) if(is_prime[i]) primes.push_back(i);
        return primes;
    }
};
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