disjoint_sets.java: Added Disjoint sets data structure rathoresrikant…

…#857

disjoint_sets.java

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+import java.io.*; 
+import java.util.*; 
+
+class DisjointUnionSets { 
+	int[] rank, parent; 
+	int n; 
+
+	// Constructor 
+	public DisjointUnionSets(int n) 
+	{ 
+		rank = new int[n]; 
+		parent = new int[n]; 
+		this.n = n; 
+		makeSet(); 
+	} 
+
+	// Creates n sets with single item in each 
+	void makeSet() 
+	{ 
+		for (int i = 0; i < n; i++) { 
+			// Initially, all elements are in 
+			// their own set. 
+			parent[i] = i; 
+		} 
+	} 
+
+	// Returns representative of x's set 
+	int find(int x) 
+	{ 
+		// Finds the representative of the set 
+		// that x is an element of 
+		if (parent[x] != x) { 
+			// if x is not the parent of itself 
+			// Then x is not the representative of 
+			// his set, 
+			parent[x] = find(parent[x]); 
+
+			// so we recursively call Find on its parent 
+			// and move i's node directly under the 
+			// representative of this set 
+		} 
+
+		return parent[x]; 
+	} 
+
+	// Unites the set that includes x and the set 
+	// that includes x 
+	void union(int x, int y) 
+	{ 
+		// Find representatives of two sets 
+		int xRoot = find(x), yRoot = find(y); 
+
+		// Elements are in the same set, no need 
+		// to unite anything. 
+		if (xRoot == yRoot) 
+			return; 
+
+		// If x's rank is less than y's rank 
+		if (rank[xRoot] < rank[yRoot]) 
+
+			// Then move x under y so that depth 
+			// of tree remains less 
+			parent[xRoot] = yRoot; 
+
+		// Else if y's rank is less than x's rank 
+		else if (rank[yRoot] < rank[xRoot]) 
+
+			// Then move y under x so that depth of 
+			// tree remains less 
+			parent[yRoot] = xRoot; 
+
+		else // if ranks are the same 
+		{ 
+			// Then move y under x (doesn't matter 
+			// which one goes where) 
+			parent[yRoot] = xRoot; 
+
+			// And increment the result tree's 
+			// rank by 1 
+			rank[xRoot] = rank[xRoot] + 1; 
+		} 
+	} 
+} 
+
+// Driver code 
+public class Main { 
+	public static void main(String[] args) 
+	{ 
+		// Let there be 5 persons with ids as 
+		// 0, 1, 2, 3 and 4 
+		int n = 5; 
+		DisjointUnionSets dus = 
+				new DisjointUnionSets(n); 
+
+		// 0 is a friend of 2 
+		dus.union(0, 2); 
+
+		// 4 is a friend of 2 
+		dus.union(4, 2); 
+
+		// 3 is a friend of 1 
+		dus.union(3, 1); 
+
+		// Check if 4 is a friend of 0 
+		if (dus.find(4) == dus.find(0)) 
+			System.out.println("Yes"); 
+		else
+			System.out.println("No"); 
+
+		// Check if 1 is a friend of 0 
+		if (dus.find(1) == dus.find(0)) 
+			System.out.println("Yes"); 
+		else
+			System.out.println("No"); 
+	} 
+}