<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 DisjointSet{
    
    public: 
    vector<int> rank, parent, size;
    // Constructor
    DisjointSet(int n){
        rank.resize(n + 1, 0);
        parent.resize(n + 1, 0);
        size.resize(n + 1, 0);
        
        // Ensures 1 Based indexing graph also
        for(int i = 0 ; i <= n ; i++){
            parent[i] = i;
            size[i] = 1;
        }
    };

    // Find Operation with path compression
    int find(int node){
        if(parent[node] == node)return node;
        else return parent[node] = find(parent[node]);
    };
    
    // Union operation by rank
    void unionbyrank(int x, int y){
        int ult_parent_x = find(x);
        int ult_parent_y = find(y);
        
        // They already belongs to the same group
        if(ult_parent_x == ult_parent_y)return;
        
        // They belong to different groups, based on ranking the union operation will be broken in three pieces
        if(rank[ult_parent_x] > rank[ult_parent_y]){
            parent[ult_parent_y] = ult_parent_x;
        }else if(rank[ult_parent_x] < rank[ult_parent_y]){
            parent[ult_parent_x] = ult_parent_y;
        }else{
            parent[ult_parent_x] = ult_parent_y;
            rank[ult_parent_y] += 1;
        }
    }
    
    // Union operation by size
    void unionbysize(int x, int y){
        int ult_parent_x = find(x);
        int ult_parent_y = find(y);
        
        // They already belongs to the same group
        if(ult_parent_x == ult_parent_y)return;
        
        // They belong to different groups, based on ranking the union operation will be broken in three pieces
        // x component bigger 
        if(size[ult_parent_x] > size[ult_parent_y]){
            parent[ult_parent_y] = ult_parent_x;
            size[ult_parent_x] += size[ult_parent_y];
        // y component bigger or both same
        }else{
            parent[ult_parent_x] = ult_parent_y;
            size[ult_parent_y] += size[ult_parent_x];
        }
    }
};
class Solution {
public:
    int largestComponentSize(vector<int>& nums) {
        int n=nums.size();
        DisjointSet ds(100001);
        for(int i=0;i<n;i++){
            for(int k=2;k<=sqrt(nums[i]);k++){
                if(nums[i]%k==0){
                    ds.unionbysize(nums[i],k);
                    ds.unionbysize(nums[i],nums[i]/k);
                }
            }
        }
        int maxi=0;
        unordered_map<int,int>mp;
        for(auto it:nums){
            mp[ds.find(it)]++;
        }
        for(int i=0;i<n;i++){
            maxi=max(maxi,mp[ds.find(nums[i])]);
        }
        return maxi;
    }
};
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