Check whether given two sets are disjoint or not

Here given code implementation process.

/*
  Java program
  Check whether given two sets are disjoint or not
*/
import java.util.HashSet;
public class SubSet
{
	//This function are displaying given array elements
	public void display(int[] arr, int size)
	{
		for (int i = 0; i < size; ++i)
		{
			System.out.print(" " + arr[i]);
		}
		System.out.print("\n");
	}
	//  Function which is determine whether element of given sets is disjoint 
	public void isDisjointSet(int[] first, int[] second)
	{
        int s1 = first.length;
        int s2 = second.length;
		// Use to check subset
		HashSet <Integer> result = new HashSet <> ();
		boolean status = true;
		for (int i = 0; i < s1; ++i)
		{
			// Get unique values of first array
			result.add(first[i]);
		}
		System.out.print(" Set A : ");
		display(first, s1);
		System.out.print(" Set B : ");
		display(second, s2);
		for (int i = 0; i < s2 && status == true; ++i)
		{
			if (result.contains(second[i]))
			{
				// When elements are already exist in resultant set A
				status = false;
			}
		}
		if (status == true)
		{
			System.out.print(" Sets are disjoint\n\n");
		}
		else
		{
			System.out.print(" Sets are not disjoint\n\n");
		}
	}
	public static void main(String[] args)
	{
		SubSet task = new SubSet();
		// Define array elements
		int[] set1 = {
			2 , 8 , 0 , 5 , 7 , 11
		};
		int[] set2 = {
			1 , 4 , 9 , 10
		};
		int[] set3 = {
			10 , 6 , 5 , 12 , 15 , 1
		};
		//Test case
		task.isDisjointSet(set1, set2);
		task.isDisjointSet(set1, set3);
	}
}

Output

 Set A :  2 8 0 5 7 11
 Set B :  1 4 9 10
 Sets are disjoint

 Set A :  2 8 0 5 7 11
 Set B :  10 6 5 12 15 1
 Sets are not disjoint
// Include header file
#include <iostream>

#include <set>

using namespace std;
class SubSet
{
	public:
		//This function are displaying given array elements
		void display(int arr[], int size)
		{
			for (int i = 0; i < size; ++i)
			{
				cout << " " << arr[i];
			}
			cout << "\n";
		}
	//  Function which is determine whether element of given sets is disjoint
	void isDisjointSet(int first[], int second[], int s1, int s2)
	{

		// Use to check subset
		set < int > result;
		bool status = true;
		for (int i = 0; i < s1; ++i)
		{
			result.insert(first[i]);
		}
		cout << " Set A : ";
		this->display(first, s1);
		cout << " Set B : ";
		this->display(second, s2);
		for (int i = 0; i < s2 && status == true; ++i)
		{
			if (result.find(second[i]) != result.end())
			{
				// When elements are already exist in resultant set A
				status = false;
			}
		}
		if (status == true)
		{
			cout << " Sets are disjoint\n\n";
		}
		else
		{
			cout << " Sets are not disjoint\n\n";
		}
	}
};
int main()
{
	SubSet task = SubSet();
	// Define array elements
	int set1[] = {
		2 , 8 , 0 , 5 , 7 , 11
	};
	int set2[] = {
		1 , 4 , 9 , 10
	};
	int set3[] = {
		10 , 6 , 5 , 12 , 15 , 1
	};
	//Get the size of arrays
	int s1 = sizeof(set1) / sizeof(set1[0]);
	int s2 = sizeof(set2) / sizeof(set2[0]);
	int s3 = sizeof(set3) / sizeof(set3[0]);
  	//Test case
	task.isDisjointSet(set1, set2, s1, s2);
	task.isDisjointSet(set1, set3, s1, s3);
	return 0;
}

Output

 Set A :  2 8 0 5 7 11
 Set B :  1 4 9 10
 Sets are disjoint

 Set A :  2 8 0 5 7 11
 Set B :  10 6 5 12 15 1
 Sets are not disjoint
// Include namespace system
using System;
using System.Collections.Generic;
/*
  C# program
  Check whether given two sets are disjoint or not
*/
public class SubSet
{
	//This function are displaying given array elements
	public void display(int[] arr, int size)
	{
		for (int i = 0; i < size; ++i)
		{
			Console.Write(" " + arr[i]);
		}
		Console.Write("\n");
	}
	//  Function which is determine whether element of given sets is disjoint
	public void isDisjointSet(int[] first, int[] second)
	{
		int s1 = first.Length;
		int s2 = second.Length;
		// Use to check subset
		HashSet < int > result = new HashSet <int> ();
		Boolean status = true;
		for (int i = 0; i < s1; ++i)
		{
			result.Add(first[i]);
		}
		Console.Write(" Set A : ");
		display(first, s1);
		Console.Write(" Set B : ");
		display(second, s2);
		for (int i = 0; i < s2 && status == true; ++i)
		{
			if (result.Contains(second[i]))
			{
				// When elements are already exist in resultant set A
				status = false;
			}
		}
		if (status == true)
		{
			Console.Write(" Sets are disjoint\n\n");
		}
		else
		{
			Console.Write(" Sets are not disjoint\n\n");
		}
	}
	public static void Main(String[] args)
	{
		SubSet task = new SubSet();
		// Define array elements
		int[] set1 = {
			2 , 8 , 0 , 5 , 7 , 11
		};
		int[] set2 = {
			1 , 4 , 9 , 10
		};
		int[] set3 = {
			10 , 6 , 5 , 12 , 15 , 1
		};
		//Test case
		task.isDisjointSet(set1, set2);
		task.isDisjointSet(set1, set3);
	}
}

Output

 Set A :  2 8 0 5 7 11
 Set B :  1 4 9 10
 Sets are disjoint

 Set A :  2 8 0 5 7 11
 Set B :  10 6 5 12 15 1
 Sets are not disjoint
<?php
/*
  Php program
  Check whether given two sets are disjoint or not
*/
class SubSet
{
	//This function are displaying given array elements
	public	function display( & $arr, $size)
	{
		for ($i = 0; $i < $size; ++$i)
		{
			echo " ". $arr[$i];
		}
		echo "\n";
	}
	//  Function which is determine whether element of given sets is disjoint
	public	function isDisjointSet( & $first, & $second)
	{
		$s1 = count($first);
		$s2 = count($second);
		// Use to check subset
		$result = array();
		$status = true;
		for ($i = 0; $i < $s1; ++$i)
		{
			if (in_array($first[$i], $result,True) == false)
            {
				// Get unique values of first array
				$result[]=$first[$i];
        	}
		}
		echo " Set A : ";
		$this->display($first, $s1);
		echo " Set B : ";
		$this->display($second, $s2);
		for ($i = 0; $i < $s2 && $status == true; ++$i)
		{
			if (in_array($second[$i], $result, True))
			{
				// When elements are already exist in resultant set A
				$status = false;
			}
		}
		if ($status == true)
		{
			echo " Sets are disjoint\n\n";
		}
		else
		{
			echo " Sets are not disjoint\n\n";
		}
	}
}

function main()
{
	$task = new SubSet();
	// Define array elements
	$set1 = array(2, 8, 0, 5, 7, 11);
	$set2 = array(1, 4, 9, 10);
	$set3 = array(10, 6, 5, 12, 15, 1);
	//Test case
	$task->isDisjointSet($set1, $set2);
	$task->isDisjointSet($set1, $set3);
}
main();

Output

 Set A :  2 8 0 5 7 11
 Set B :  1 4 9 10
 Sets are disjoint

 Set A :  2 8 0 5 7 11
 Set B :  10 6 5 12 15 1
 Sets are not disjoint
/*
  Node Js program
  Check whether given two sets are disjoint or not
*/
class SubSet
{
	//This function are displaying given array elements
	display(arr, size)
	{
		for (var i = 0; i < size; ++i)
		{
			process.stdout.write(" " + arr[i]);
		}
		process.stdout.write("\n");
	}
	//  Function which is determine whether element of given sets is disjoint
	isDisjointSet(first, second)
	{
		var s1 = first.length;
		var s2 = second.length;
		// Use to check subset
		var result = new Set();
		var status = true;
		for (var i = 0; i < s1; ++i)
		{
			result.add(first[i]);
		}
		process.stdout.write(" Set A : ");
		this.display(first, s1);
		process.stdout.write(" Set B : ");
		this.display(second, s2);
		for (var i = 0; i < s2 && status == true; ++i)
		{
			if (result.has(second[i]))
			{
				// When elements are already exist in resultant set A
				status = false;
			}
		}
		if (status == true)
		{
			process.stdout.write(" Sets are disjoint\n\n");
		}
		else
		{
			process.stdout.write(" Sets are not disjoint\n\n");
		}
	}
}

function main()
{
	var task = new SubSet();
	// Define array elements
	var set1 = [2, 8, 0, 5, 7, 11];
	var set2 = [1, 4, 9, 10];
	var set3 = [10, 6, 5, 12, 15, 1];
	//Test case
	task.isDisjointSet(set1, set2);
	task.isDisjointSet(set1, set3);
}
main();

Output

 Set A :  2 8 0 5 7 11
 Set B :  1 4 9 10
 Sets are disjoint

 Set A :  2 8 0 5 7 11
 Set B :  10 6 5 12 15 1
 Sets are not disjoint
#   Python 3 program
#   Check whether given two sets are disjoint or not

class SubSet :
	# This function are displaying given list elements
	def display(self, arr, size) :
		i = 0
		while (i < size) :
			print(" ", arr[i], end = "")
			i += 1
		
		print(end = "\n")
	
	#   Function which is determine whether element of given sets is disjoint
	def isDisjointSet(self, first, second) :
		s1 = len(first)
		s2 = len(second)
		#  Use to check subset
		result = set()
		status = True
		i = 0
		while (i < s1) :
			#  Get unique values of first list
			result.add(first[i])
			i += 1
		
		print(" Set A : ", end = "")
		self.display(first, s1)
		print(" Set B : ", end = "")
		self.display(second, s2)
		i = 0
		while (i < s2 and status == True) :
			if (second[i] in result) :
				#  When elements are already exist in resultant set A
				status = False
			
			i += 1
		
		if (status == True) :
			print(" Sets are disjoint\n")
		else :
			print(" Sets are not disjoint\n")
		
	

def main() :
	task = SubSet()
	#  Define list elements
	set1 = [2, 8, 0, 5, 7, 11]
	set2 = [1, 4, 9, 10]
	set3 = [10, 6, 5, 12, 15, 1]
	# Test case
	task.isDisjointSet(set1, set2)
	task.isDisjointSet(set1, set3)

if __name__ == "__main__": main()

Output

 Set A :   2  8  0  5  7  11
 Set B :   1  4  9  10
 Sets are disjoint

 Set A :   2  8  0  5  7  11
 Set B :   10  6  5  12  15  1
 Sets are not disjoint
#   Ruby program
#   Check whether given two sets are disjoint or not

class SubSet 
	# This function are displaying given array elements
	def display(arr, size) 
		i = 0
		while (i < size) 
			print(" ", arr[i])
			i += 1
		end

		print("\n")
	end

	#   Function which is determine whether element of given sets is disjoint
	def isDisjointSet(first, second) 
		s1 = first.length
		s2 = second.length
		#  Use to check subset
		result = []
		status = true
		i = 0
		while (i < s1) 
			result.push(first[i])
			i += 1
		end

		print(" Set A : ")
		self.display(first, s1)
		print(" Set B : ")
		self.display(second, s2)
		i = 0
		while (i < s2 && status == true) 
			if (result.include?(second[i])) 
				#  When elements are already exist in resultant set A
				status = false
			end
			i += 1
		end

		if (status == true) 
			print(" Sets are disjoint\n\n")
		else 
			print(" Sets are not disjoint\n\n")
		end

	end

end

def main() 
	task = SubSet.new()
	#  Define array elements
	set1 = [2, 8, 0, 5, 7, 11]
	set2 = [1, 4, 9, 10]
	set3 = [10, 6, 5, 12, 15, 1]
	# Test case
	task.isDisjointSet(set1, set2)
	task.isDisjointSet(set1, set3)
end

main()

Output

 Set A :  2 8 0 5 7 11
 Set B :  1 4 9 10
 Sets are disjoint

 Set A :  2 8 0 5 7 11
 Set B :  10 6 5 12 15 1
 Sets are not disjoint

import scala.collection.mutable._;
/*
  Scala program
  Check whether given two sets are disjoint or not
*/
class SubSet
{
	//This function are displaying given array elements
	def display(arr: Array[Int], size: Int): Unit = {
		var i: Int = 0;
		while (i < size)
		{
			print(" " + arr(i));
			i += 1;
		}
		print("\n");
	}
	//  Function which is determine whether element of given sets is disjoint
	def isDisjointSet(first: Array[Int], second: Array[Int]): Unit = {
		var s1: Int = first.length;
		var s2: Int = second.length;
		// Use to check subset
		var result: Set[Int] = Set();
		var status: Boolean = true;
		var i: Int = 0;
		while (i < s1)
		{
			result.add(first(i));
			i += 1;
		}
		print(" Set A : ");
		this.display(first, s1);
		print(" Set B : ");
		this.display(second, s2);
		i = 0;
		while (i < s2 && status == true)
		{
			if (result.contains(second(i)))
			{
				// When elements are already exist in resultant set A
				status = false;
			}
			i += 1;
		}
		if (status == true)
		{
			print(" Sets are disjoint\n\n");
		}
		else
		{
			print(" Sets are not disjoint\n\n");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: SubSet = new SubSet();
		// Define array elements
		var set1: Array[Int] = Array(2, 8, 0, 5, 7, 11);
		var set2: Array[Int] = Array(1, 4, 9, 10);
		var set3: Array[Int] = Array(10, 6, 5, 12, 15, 1);
		//Test case
		task.isDisjointSet(set1, set2);
		task.isDisjointSet(set1, set3);
	}
}

Output

 Set A :  2 8 0 5 7 11
 Set B :  1 4 9 10
 Sets are disjoint

 Set A :  2 8 0 5 7 11
 Set B :  10 6 5 12 15 1
 Sets are not disjoint
/*
  Swift 4 program
  Check whether given two sets are disjoint or not
*/
class SubSet
{
	//This function are displaying given array elements
	func display(_ arr: [Int], _ size: Int)
	{
		var i: Int = 0;
		while (i < size)
		{
			print(" ", arr[i], terminator: "");
			i += 1;
		}
		print(terminator: "\n");
	}
	//  Function which is determine whether element of given sets is disjoint
	func isDisjointSet(_ first: [Int], _ second: [Int])
	{
		let s1: Int = first.count;
		let s2: Int = second.count;
		// Use to check subset
		var result = Set<Int>()
		var status: Bool = true;
		var i: Int = 0;
		while (i < s1)
		{
			result.insert(first[i]);
			i += 1;
		}
		print(" Set A : ", terminator: "");
		self.display(first, s1);
		print(" Set B : ", terminator: "");
		self.display(second, s2);
		i = 0;
		while (i < s2 && status == true)
		{
			if (result.contains(second[i]))
			{
				// When elements are already exist in resultant set A
				status = false;
			}
			i += 1;
		}
		if (status == true)
		{
			print(" Sets are disjoint\n");
		}
		else
		{
			print(" Sets are not disjoint\n");
		}
	}
}
func main()
{
	let task: SubSet = SubSet();
	// Define array elements
	let set1: [Int] = [2, 8, 0, 5, 7, 11];
	let set2: [Int] = [1, 4, 9, 10];
	let set3: [Int] = [10, 6, 5, 12, 15, 1];
	//Test case
	task.isDisjointSet(set1, set2);
	task.isDisjointSet(set1, set3);
}
main();

Output

 Set A :   2  8  0  5  7  11
 Set B :   1  4  9  10
 Sets are disjoint

 Set A :   2  8  0  5  7  11
 Set B :   10  6  5  12  15  1
 Sets are not disjoint
/*
  Kotlin program
  Check whether given two sets are disjoint or not
*/
class SubSet
{
	//This function are displaying given array elements
	fun display(arr: Array < Int > , size: Int): Unit
	{
		var i: Int = 0;
		while (i < size)
		{
			print(" " + arr[i]);
			i += 1;
		}
		print("\n");
	}
	//  Function which is determine whether element of given sets is disjoint
	fun isDisjointSet(first: Array <Int> , second: Array <Int> ): Unit
	{
		var s1: Int = first.count();
		var s2: Int = second.count();
		// Use to check subset
		var result: MutableSet <Int> = mutableSetOf <Int> ();
		var status: Boolean = true;
		var i: Int = 0;
		while (i < s1)
		{
			result.add(first[i]);
			i += 1;
		}
		print(" Set A : ");
		this.display(first, s1);
		print(" Set B : ");
		this.display(second, s2);
		i = 0;
		while (i < s2 && status == true)
		{
			if (result.contains(second[i]))
			{
				// When elements are already exist in resultant set A
				status = false;
			}
			i += 1;
		}
		if (status == true)
		{
			print(" Sets are disjoint\n\n");
		}
		else
		{
			print(" Sets are not disjoint\n\n");
		}
	}
}
fun main(args: Array < String > ): Unit
{
	var task: SubSet = SubSet();
	// Define array elements
	var set1: Array < Int > = arrayOf(2, 8, 0, 5, 7, 11);
	var set2: Array < Int > = arrayOf(1, 4, 9, 10);
	var set3: Array < Int > = arrayOf(10, 6, 5, 12, 15, 1);
	//Test case
	task.isDisjointSet(set1, set2);
	task.isDisjointSet(set1, set3);
}

Output

 Set A :  2 8 0 5 7 11
 Set B :  1 4 9 10
 Sets are disjoint

 Set A :  2 8 0 5 7 11
 Set B :  10 6 5 12 15 1
 Sets are not disjoint


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