<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 ds {
    public:
        vector<int> parent;
        vector<int> rank;
        ds(int n) {
            parent.resize(n, 0);
            rank.resize(n, 0);

            for (int i = 0; i < n; i++) 
                parent[i] = i;
        }

        int find(int node) {
            return parent[node] = (parent[node] == node) ? node: find(parent[node]);
        }

        void reset(int node) {
            parent[node] = node;
        }

        void Union(int nodeA, int nodeB) {
            int uParA = find(nodeA), uParB = find(nodeB);
            if (uParA != uParB) {
                rank[uParA] < rank[uParB] ? parent[uParA] = uParB : parent[uParB] = uParA;
                rank[uParA] += rank[uParA] == rank[uParB] ? 1: 0;
            }
        }
};

const int MAX = 1e5 + 5;
bool prime[MAX];
int spf[MAX];

class Solution {
public:
    void preprocess() {
        fill(prime, prime + MAX, true);
        for (int i = 1; i < MAX; i++) {
            spf[i] = i;
        }
        for (int i = 2; i * i < MAX; i++) {
            if (prime[i]) {
                for (int j = i * i; j < MAX; j += i) {
                    if (prime[j])
                        spf[j] = i;
                    prime[j] = false;
                }
            }
        }
    }

    unordered_set<int> getFactors(int x) {
            unordered_set<int> res;
            while (x > 1) {
                res.insert(spf[x]);
                x /= spf[x];
            }
            return res;
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        preprocess();
        unordered_map<int, vector<int>> rmp;
        for (int i = 0; i < n; i++) {
            unordered_set<int> st = getFactors(nums[i]);
            for (auto ele : st) {
                rmp[ele].push_back(i);
            }
        }
        ds d(n);
        for (auto &p : rmp) {
            vector<int> temp = p.second;
            if (temp.size() <= 1)
                continue;
            for (int i = 1; i < temp.size(); i++) {
                d.Union(temp[i], temp[i-1]);
            }
        }
        for (int i = 1; i < n; i++) {
            if (d.find(i) != d.find(i-1)) {
                return false;
            }
        }
        return true;
    }
};
234962