<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:
    void bfs(int root, vector<bool> & vis, vector<vector<int>> & adj) {
        queue<int> q; 
        q.push(root); 
        vis[root] = true; 

        while (!q.empty()) {
            auto u = q.front(); 
            q.pop();             

            for (auto v : adj[u]) {
                if (!vis[v]) {
                    vis[v] = true; 
                    q.push(v); 
                }
            }
        }

        return; 
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size(); 
        unordered_map<int, vector<int>> mp(317); 

        for (int i = 0; i < n; i++) {
            int num = nums[i];
            for (int div = 2; div * div <= num; div++) {
                if (num % div == 0) {
                    mp[div].push_back(i);
                    while (num % div == 0) {
                        num /= div;
                    }
                }
            }

            if (num > 1) 
                mp[num].push_back(i);
        }

        vector<vector<int>> adj(n); 

        for (auto & [key, e] : mp) {
            for (int j = 1; j < e.size(); j++) {
                int u = e[j], v = e[j - 1]; 
                adj[u].push_back(v); 
                adj[v].push_back(u); 
            }
        }

       vector<bool> vis(n, false); 

        bfs(0, vis, adj); 
        bool ans = true; 

        for (auto flg : vis) 
            ans &= flg; 

        return ans; 
    }
};
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