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

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

    void join(int x,int y){
        int rootX = find(x);
        int rootY = find(y);

        if(rootX != rootY){
            if(rank[rootY] < rank[rootX]){
                parent[rootY] = rootX;
            } 
            else if(rank[rootY] > rank[rootX]){
                parent[rootX] = rootY;
            }
            else{
                parent[rootY] = rootX;
                rank[rootX]++;
            }
        }
    }
};

static const int N = 1e5 + 100;

vector<int> mp[N];

bool res = false;

void get(int n, unordered_set<int>& factors) {
    while (n % 2 == 0) {
        factors.insert(2);
        n /= 2;
    }
    for (int i = 3; i <= sqrt(n); i += 2) {
        while (n % i == 0) {
            factors.insert(i);
            n /= i;
        }
    }

    if (n > 2) {
        factors.insert(n);
    }
}

void precalc(){
    for(int i = 1;i < N - 60;i++){
        unordered_set<int> fac;
        get(i,fac);
        for(auto j : fac){
            mp[i].push_back(j);
        }
    }
}

class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        unordered_set<int> st;
        if(nums.size() == 1){
            return true;
        }
        for(int i : nums){
            st.insert(i);
        }
        if(st.find(1) != st.end()){
            return false;
        }
        int n = st.size();
        nums.clear();
        for(auto i : st){
            nums.push_back(i);
        }

        if(!res){
            precalc();
            res = true;
        }
        DSU dsu(n);

        unordered_map<int,int> indexStore;

        for(int i = 0;i < n;i++){
            vector<int> primes = mp[nums[i]];

            for(auto j : primes){
                if(indexStore.find(j) != indexStore.end()){
                    dsu.join(indexStore[j],i);
                }
                else{
                    indexStore[j] = i;
                }
            }
        }

        unordered_map<int,int> comps;

        for(int i = 0;i < n;i++){
            int par = dsu.find(i);
            comps[par]++;
        }

        return comps.size() == 1;


    }
};
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