<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>

typedef long long ll;

class Solution {
public:
    vector<int> prime_factor;
    void init(int MAX){
        prime_factor.resize(MAX+1, 1);
        for (int i=2;i<=MAX;i++){
            if (prime_factor[i] != 1)continue;
            // prime number
            prime_factor[i] = i;
            for (ll j = (ll)i*(ll)i;j<=MAX;j+=i)prime_factor[j]=i;
        }
    }
    
    vector<int> get_prime_factor(int n){
        vector<int> factor;
        while (n > 1){
            factor.push_back(prime_factor[n]);
            n/=prime_factor[n];
        }
        return factor;
    }

    vector<int> parent;
    vector<int> size;

    int find_set(int u){
        if (u == parent[u]) return u;
        return parent[u] = find_set(parent[u]);
    }

    void union_set(int u, int v){
        u = find_set(u);
        v = find_set(v);
        if (u != v){
            if (size[u] > size[v]){
                parent[v] = u;
                size[u]+=size[v];
            }else {
                parent[u] = v;
                size[v]+=size[u];
            }
        }
    }
    bool canTraverseAllPairs(vector<int>& nums) {
        int MAX{INT_MIN};
        for (const int& num: nums)MAX=max(MAX, num);
        init(MAX);
        // prime -> [index having factor in it]
        unordered_map<int, vector<int>> table;
        for (int i=0;i<nums.size();i++){
            int num = nums[i];
            auto factors = get_prime_factor(num);
            for (const int& f: factors)table[f].push_back(i);
        }

        parent.resize(nums.size());
        size.resize(nums.size());
        for (int i=0;i<nums.size();i++){
            parent[i]=i;
            size[i] = 1;
        }

        for (auto itr: table){
            auto idx = itr.second;
            for (int i=1;i<idx.size();i++){
                union_set(idx[i-1], idx[i]);
            }
        }
        int leader = find_set(0);
        for (int i=1;i<nums.size();i++){
            if (find_set(i) != leader) return false;
        }
        return true;
    }
};
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