<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:
    vector<int> parent;
    vector<int> rank;

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

    void Union(int x, int y) {
        int x1 = find(x);
        int y1 = find(y);
        if (x1 == y1) {
            return;
        }
        if (rank[x1] > rank[y1]) {
            parent[y1] = x1;
        } else if (rank[y1] > rank[x1]) {
            parent[x1] = y1;
        } else {
            parent[x1] = y1;
            rank[y1]++;
        }
    }

    void Sieve(int n, const vector<int>& nums, unordered_set<int>& unique_nums) {
        vector<bool> prime(n + 1, true);
        for (int i = 2; i <= n; ++i) {
            if (prime[i]) {
                int prev = i;
                for (int j = i; j <= n; j += i) {
                    prime[j] = false;
                    if (unique_nums.count(j)) {
                        if (prev != -1) {
                            Union(prev, j);
                        }
                       
                    }
                }
            }
        }
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        int max_num = *max_element(nums.begin(), nums.end());
        if(n==1)
        {
            return true ;
        }

        parent.resize(max_num + 1);
        rank.resize(max_num + 1, 0);
        for (int i = 0; i <= max_num; ++i) {
            parent[i] = i;
        }

        unordered_set<int> unique_nums(nums.begin(), nums.end());
        Sieve(max_num, nums, unique_nums);

        int root = find(nums[0]);
        for (const int& num : nums) {
            if(num==1)
            {
                return false ;
            }
            if (find(num) != root) {
                return false;
            }
        }
        return true;
    }
};

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