<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:  
    // const int MAXN = 100001;
    vector<vector<int>> prime_factors;

     void prime (int MAXN){
        prime_factors.resize(MAXN + 1);
    
         for (int i = 2; i <= MAXN; i++) {
            if (prime_factors[i].empty()) 
            { 
                for (int j = i; j <= MAXN; j += i) {
                    prime_factors[j].push_back(i); 
                }
            }
        }
    }

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

        int findUparent(int x) {
            if (x == parent[x]) return x;
            return parent[x] = findUparent(parent[x]);  
        }

        void unionBySize(int u, int v) {
            int ult_u = findUparent(u);
            int ult_v = findUparent(v);
            if (ult_u == ult_v) 
            {
                return;
            }
            if (size[ult_u] > size[ult_v]) {
                size[ult_u] += size[ult_v];
                parent[ult_v] = ult_u;
            } else {
                parent[ult_u] = ult_v;
                size[ult_v] += size[ult_u];
            }
        }

        int sizeOf(int x) {
            return size[findUparent(x)];
        }
        void resize(int par, int val)
        {
            size[par]-= val;
        }
    };

    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        if(n==1)
        return true;
        int maxi = *max_element(nums.begin(),nums.end());
        int mini = *(min_element(nums.begin(),nums.end()));
        prime(maxi);
        if(mini == 1)
        return 0;
        sort(nums.begin(),nums.end());
        DisjointSet ds(maxi);
        vector<int> vis(maxi+1,1);
        for (int i = 0; i < n; i++) {
            int num = nums[i];
            int siz = prime_factors[num].size();
            vis[num] = 0;
            for (int j = 0; j < siz; j++) {
                ds.unionBySize(num,prime_factors[num][j]);
            }
            int par = ds.findUparent(num);
            for(int j =0;j<siz;j++)
            {
                ds.resize(par,vis[prime_factors[num][j]]);
                vis[prime_factors[num][j]] = 0;
            }
        }
        set<int> st;
        for(int i=0;i<n;i++)
        st.insert(nums[i]);
        int unique = st.size();
        bool flag = false;
        for (int i = 1; i <= maxi; i++) {
            if (ds.sizeOf(i) >= unique) {  
                flag = true;
                break;
            }
        }
        return flag;
    }
};

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