<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 {

    vector<int> parent;

    vector<int> size;

public:

    DSU(int N) {

        for (int i = 0; i <= N; i++) {

            parent.push_back(i);

            size.push_back(1);

        }

    }

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

    void merge(int x, int y) {

        int fx = find(x);

        int fy = find(y);

        if (fx == fy) return;

        if (size[fx] > size[fy]) swap(fx, fy);

        parent[fx] = fy;

        size[fy] += size[fx];

    }

};

class Solution {

public:

    bool canTraverseAllPairs(vector<int>& nums) {

        int MAX = 100000;

        int N = nums.size();

        vector<bool> has(MAX + 1, false);

        for (int x : nums) {

            has[x] = true;

        }

        if (N == 1) return true;

        if (has[1]) return false;

        vector<int> sieve(MAX + 1, 0);

        for (int d = 2; d <= MAX; d++) {

            if (sieve[d] == 0) {

                for (int v = d; v <= MAX; v += d) {

                    sieve[v] = d;

                }

            }

        }

        DSU unionfind(2 * MAX + 1);

        for (int x : nums) {

            int val = x;

            while (val > 1) {

                int prime = sieve[val];

                int root = prime + MAX;

                if (unionfind.find(root) != unionfind.find(x)) {

                    unionfind.merge(root, x);

                }

                while (val % prime == 0) {

                    val /= prime;

                }

            }

        }

        int cnt = 0;

        for (int i = 2; i <= MAX; i++) {

            if (has[i] && unionfind.find(i) == i) {

                cnt++;

            }

        }

        return cnt == 1;

    }

};



    

        

    


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