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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