<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:
    unordered_map<int, vector<int>> prime2index;
    unordered_map<int, vector<int>> index2prime;
    void dfs(int index, vector<int>& visitedIndex, unordered_map<int,bool>& visitedPrime){
        if(visitedIndex[index] == true) return;
        visitedIndex[index] = true;

        for(auto &prime : index2prime[index]){
            if(visitedPrime[prime] == true) 
                continue;
            visitedPrime[prime] = true;
            for(auto &index1 : prime2index[prime]){
                if(visitedIndex[index1] == true) continue;
                dfs(index1, visitedIndex, visitedPrime);
            }
        }
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        for (int i=0; i<nums.size(); i++) {
            int temp = nums[i];
            for (int j = 2; j*j <= nums[i]; j++) {
                if (temp % j == 0) {
                    prime2index[j].push_back(i);
                    index2prime[i].push_back(j);
                    while (temp % j == 0)
                      temp /= j;
                }
            }
            if (temp > 1) {
                prime2index[temp].push_back(i);
                index2prime[i].push_back(temp);
            }
        }

        vector<int> visitedIndex(nums.size(),0);
        unordered_map<int,bool> visitedPrime;
        dfs(0, visitedIndex, visitedPrime);     

        for(int i=0; i<visitedIndex.size(); i++) 
            if(visitedIndex[i] == false) 
                return false;
        return true;    
    }
};




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