<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 DisjointSet{
    public:
    vector<int>size ,parent;
        DisjointSet(int n){
            size.resize(n+1,1);
            parent.resize(n+1,0);
            for(int i=0;i<n+1;i++){
                parent[i] =i;
            }
        }

        int find_parent(int node){
            if(node==parent[node]){
                return node;
            }

            int call = find_parent(parent[node]);

            return parent[node] =call;
        }

        void UnionBySize(int u , int v){
            int ultimate_par_u = find_parent(u);
            int ultimate_par_v = find_parent(v);
            if(ultimate_par_u==ultimate_par_v) return;

            if(size[ultimate_par_u]<size[ultimate_par_v]){
                size[ultimate_par_v]+=size[ultimate_par_u];
                parent[ultimate_par_u]=ultimate_par_v;

            }
            else {
                size[ultimate_par_v]+=size[ultimate_par_u];
                parent[ultimate_par_v]=ultimate_par_u;

            }
        }   

};

class Solution {
public:
    void pf(int number, vector<int>& arr) {
        for (int i = 2; i * i <= number; i++) {
            if (number % i == 0) {
                arr.push_back(i);
                while (number % i == 0) {
                    number /= i;
                }
            }
        }
        if (number > 1) {
            arr.push_back(number);
        }
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        if(n <= 1) return true;

        // Check if there's a 1 in the array
        // for(int num : nums) {
        //     if(num == 1) return n == 1;
        // }

        unordered_map<int, vector<int>> mp;

        for(int i = 0; i < n; i++) {
            int number = nums[i];
            vector<int> prim_factors;
            pf(number, prim_factors);
            for(int factor : prim_factors) {
                mp[factor].push_back(i);
            }
        }

        DisjointSet dsu(n);
        
        for(auto& entry : mp) {
            const vector<int>& indices = entry.second;
            for(int i = 1; i < indices.size(); i++) {
                dsu.UnionBySize(indices[i], indices[i-1]);
            }
        }

        int representative = dsu.find_parent(0);
        for(int i = 0; i < n; i++) {
            if(dsu.find_parent(i) != representative) {
                return false;
            }
        }

        return true;
    }
};
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