<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 DSU{
    public:
    vector<int>par,size;
    DSU(int n){
        par.resize(n+1);
        size.resize(n+1);
        for(int i=0;i<=n;i++){
            par[i] = i;
            size[i] = 1;
        }
    }
        int fpar(int node){
            if(par[node]!=node)return par[node] = fpar(par[node]);
            return node;
        }
        void unionn(int u,int v){
            int ul_u = fpar(u);
            int ul_v = fpar(v);
            if(ul_u == ul_v)return;
            if(size[ul_u] > size[ul_v]){
                par[ul_v] = ul_u;
                size[ul_u]+=size[ul_v];
            }
            else{
                par[ul_u] = ul_v;
                size[ul_v]+=size[ul_u];
            }
        }
};
class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        if(n == 1)return true;
        int m = 100000;
        bool prime[m+1];
        memset(prime,true,sizeof(prime));
        prime[1] = false;
        for(int p = 2;p*p <= m;p++){
            if(prime[p]){
                for(int i = p*p;i<=m;i+=p){
                    prime[i] = false;
                }
            }
        }
        DSU ds(n);
        unordered_map<int,vector<int>>mp;
        for(int i=0;i<n;i++){
            for(int d=2;d*d<=nums[i];d++){
                if(nums[i]%d == 0){
                    int v = nums[i]/d;
                    if(prime[d]){
                        mp[d].push_back(i);
                    }
                    if(prime[v]){
                        mp[v].push_back(i);
                    }
                }
            }
            if(prime[nums[i]]){
                mp[nums[i]].push_back(i);
            }
        }
        for(auto it:mp){
            vector<int>a = it.second;
            for(int i = 1;i<a.size();i++){
                ds.unionn(a[0],a[i]);
            }
        }
        set<int>comp;
        for(int i=0;i<n;i++){
            if(nums[i] == 1)return false;
            comp.insert(ds.fpar(i));
        }
        return comp.size() == 1;
    }
};
126062