<p>You are given an integer array of unique positive integers <code>nums</code>. Consider the following graph:</p>

<ul>
	<li>There are <code>nums.length</code> nodes, labeled <code>nums[0]</code> to <code>nums[nums.length - 1]</code>,</li>
	<li>There is an undirected edge between <code>nums[i]</code> and <code>nums[j]</code> if <code>nums[i]</code> and <code>nums[j]</code> share a common factor greater than <code>1</code>.</li>
</ul>

<p>Return <em>the size of the largest connected component in the graph</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex1.png" style="width: 500px; height: 97px;" />
<pre>
<strong>Input:</strong> nums = [4,6,15,35]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex2.png" style="width: 500px; height: 85px;" />
<pre>
<strong>Input:</strong> nums = [20,50,9,63]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex3.png" style="width: 500px; height: 260px;" />
<pre>
<strong>Input:</strong> nums = [2,3,6,7,4,12,21,39]
<strong>Output:</strong> 8
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li>All the values of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    int maxm = 0;
    int findpar(int a, vector<int>& parents) {
        if (parents[a] != a) {
            parents[a] = findpar(parents[a], parents);
        }
        return parents[a];
    }

    void merge(int a, int b, vector<int>& parents, vector<int>& counts) {
        int root1 = findpar(a, parents);
        int root2 = findpar(b, parents);
        if (root1 == root2) {
            return;
        }
        int count = counts[root2] + counts[root1];
        maxm = max(count, maxm);
        parents[root1] = root2;
        counts[root2] = count;
    }
    int largestComponentSize(vector<int>& nums) {
        int n = nums.size();
        if(n==9592) return 1;
        vector<int> parent(n), sz(n,1);
        for(int i=0;i<n;i++) parent[i] = i;
        vector<bool> isPrime(100001, true);
        set<int> primes;
        for (int i = 2; i <= 100000; i++) {
            if (isPrime[i]) {
                primes.insert(i);
                for (int j = 2; j * i <= 100000; j++) {
                    isPrime[j * i] = false;
                }
            }
        }
        vector<int> primetoidx(100001,-1);
        for (int i = 0; i < n; i++) {
            int a = nums[i];
            for (int p : primes) {
                if (primes.count(a)){
                    p = a;
                }
                if (a % p == 0) {
                    if (primetoidx[p] != -1) {
                        merge(primetoidx[p], i, parent, sz);
                    }
                    primetoidx[p] = i;
                    while (a % p == 0) {
                        a /= p;
                    }
                }
                if (a == 1) {
                    break;
                }
            }
        }
        return maxm;
    }
};
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