Disjoint Set Graph Learning

Disjoint set is a data structures used in Kruskal minimal spanning tree.
This data structures allows us to create the union of two or more nodes.
It lets us determine if two nodes belong to the same component of the graph of not.
Time complexity is O(4alpha) (if we are using path compression which we are, else it will be logarithmic) which is constant time complexity that has been proven.

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class Main{
    int parent[]  = new int[100000];
    int rank[] = new int[100000];
    void makeSet(){
        for(int i=1;i<=n;i++){
            parent[i] = i; // initially parent of node i is i itself
            rank[i] =0;// intially rank of all the nodes are 0
        }
    }

    int findParent(int node){
        if(node == parent[node]) return node; // ie if the parent of 'node' is 'node' itself then just return the node.
        //return findParent(parent[node]);// else recursively call findParent to get the parent of this union.
        // in order to path conpress (making sure that every node is connected to its parent in the given union)
        return parent[node]  = findParent(parent[node]);
        //example : 7->6->4->3 , here 3 is the parent of this union, so in order to get the parent of 7 which is 3 we can path compress it. like 7->3,6->3,4->3 etc.
    }
    void union(int u, int v){
        u = findParent(u);
        v = findParent(v);
        if(rank[u] < rank[v]) parent[u] =v;
        else if (rank[u] > rank[v]) parent[v] = u;
        else parent[u] =v; // or parent[v] = u;
        // here u and v had the same rank
        //and now v became parent of u hence its rank will increase
        rank[v]++;
    }

    public static void main(String args[]) throws Exception{
        Main m = new Main();
        //if we where given no of nodes as n
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        while(n>0){
            int u = n--;
            int v = n--;
            m.union(u,v);// this will create union of n nodes
        }
        // if 2 and 3 belong to the same component or not find out
        if(findParent(2)! = findParent(3)) System.out.println("Different component");
        else System.out.println("Same component");
    }
}

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