<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>

///////////////////////////////////////////////////

#include <bits/stdc++.h>

using namespace std;

#define INT(x) int x; cin >> x;
#define VECTOR(v) vector<int> v(n); for (int i = 0; i < n; i ++) { cin >> v[i]; }
#define EXISTS(s, x) s.find(x) != s.end()
#define ALL(v) v.begin(),v.end()
#define VI vector<int>

///////////////////////////////////////////////////


struct UnionFind {
	vector<int> par;
	vector<int> rank;

	UnionFind(int n) {
		par = vector(n, 0);
		rank = vector(n, 1);

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

	int root(int x) {
		while (x != par[x]) x = par[x];
		
		return x;
	}

	void join(int x, int y) {
		int p1 = root(x), p2 = root(y);

		if (p1 == p2) return;
		if (rank[p2] > rank[p1]) swap(p1, p2);
		
		par[p2] = par[p1];
		rank[p1] += rank[p2];
	}
};

class Solution {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();

        UnionFind uf(n);

        unordered_map<int, int> lookup;
        for (int i = 0; i < n; i ++) {
            if (EXISTS(lookup, nums[i]) && nums[i] != 1) {
                uf.join(lookup[nums[i]], i);
            }

            lookup[nums[i]] = i;
        }

        int _max = *max_element(ALL(nums));
        for (int x = 2; x <= _max; x ++ ) {
            int prev = -1;
            if (EXISTS(lookup, x)) prev = lookup[x];

            for (int k = 2 * x; k <= _max; k += x) {
                if (EXISTS(lookup, k)) {
                    if (prev == -1) prev = lookup[k];

                    uf.join(prev, lookup[k]);
                }
            }
        }

        int r = uf.root(0);
        for (int i = 0; i < n; i ++) {
            if (uf.root(i) != r) return false;
        }

        return true;
    }
};
101862