<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 union_find {
public:
    vector<int> p;
    vector<int> r;
    union_find(int n) {
        p.resize(n, 0);
        r.resize(n, 0);
        
        for(int i = 0 ; i < n ; i++) {
            p[i] = i;
        }
    }
    
    int Find(int x) {
        if(p[x] != x) {
            p[x] = Find(p[x]);
        }
        return p[x];
    }
    
    void Union(int x, int y) {
        int px = Find(x);
        int py = Find(y);
        if(px == py) {
            return;
        }
        
        if(r[px] > r[py]) {
            p[py] = px;
        }
        else if(r[py] > r[px]) {
            p[px] = py;
        }
        else {
            p[py] = px;
            r[px]++;
        }
    }
    
};
class Solution {
public:
    void find_factors(int val, vector<int> &v) {
        for(int j = 2 ; (j * j) <= val ; j++) {
            bool add = false;
            while((val % j) == 0) {
                val = val / j;
                add = true;
            }
            if(add) {
                v.push_back(j);
            }
        }

        if(val > 1) {
            v.push_back(val);
        }
    }
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> v(n);
        union_find uf(100000);
        unordered_set<int> st;
        
        if(n == 1) {
            return true;
        }
        
        //check failed case 
        for(int i = 0 ; i < n; i++) {
            if(nums[i] == 1) {
                return false;
            }
            find_factors(nums[i], v[i]);
        }
        for(int i = 0 ; i < n ; i++) {
            for(int j = 1 ; j < v[i].size(); j++) {
                uf.Union(v[i][0], v[i][j]);
            }
        }
        
        for(int i = 0 ; i < n ; i++) {
            for(int j = 0 ; j < v[i].size(); j++) {
                st.insert(uf.Find(v[i][j]));
            }
        }
        
        
        return st.size() == 1;
    }
};
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