Minimum Spanning Tree (Kruskal's algorithm) Using Disjoint set
Prashant Mishra
Posted on August 16, 2022
Time complexity is O(mlogm) + (m*O(4alpha)
~ constant time = O(1)
for disjoint set )
Hence,
Efficient time complexity is : O(mlogm)
where m
is the no of edges in the graph logm
is the time complexity to sort the List of m
edges.
For more information go through
class Main{
int findParent(int node,int parent[]){
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 compress (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,int parent[], int rank[]){
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{
int n =5;
ArrayList<Node> adj = new ArrayList<>();
adj.add(new Node(0,1,2));
adj.add(new Node(0,3,6));
adj.add(new Node(1, 3, 8));
adj.add(new Node(1, 2, 3));
adj.add(new Node(1, 4, 5));
adj.add(new Node(2, 4, 7));
Main obj = new Main();
obj.kruskalAlgo(adj,n);
}
public void kruskalAlgo(ArrayList<Node> adj, int n){
Collections.sort(adj, new Comparator<Node> {
public int compare(Node a, Node b){
return a.getW()-b.getW();// sorting in ascending order of the weight;
}
});
int parent[] = new int[n];
int rank[] = new int[n];
//makeSet();
for(int i =0;i<n;i++){
parent[i] =i;
rank[i] = 0;
}
int costOfMst = 0;
ArrayList<Node> mst = new ArrayList<>();
for(Node it : adj){
// taking the node that is not taken already to avoid cycle.
if(findParent(it.getU(),parent)!=findParent(it.getV(),parent)){
costOfMst+=it.getW(); // add the cost in the costOfMst to keep track of growing cost to cover all the nodes
mst.add(it);
union(it.getU(),it.getV(),parent,rank); // make the node part of the union
}
}
System.out.println(costOfMst);
for(Node it: mst){
System.out.println(it.getU()+" - "+ it.getV());
}
}
class Node {
int u;
int v;
int weight;
public Node(int u,int v, int w){
this.u = u;
this.v = v;
this.weight = w;
}
public int getU(){
return this.u;
}
public int getV(){
return this.v;
}
public int getW(){
return this.weight;
}
}
}
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Prashant Mishra
Posted on August 16, 2022
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