How to check if two given sets are disjoint?

Given two sets represented by two arrays, how to check if the given two sets are disjoint or not? It may be assumed that the given arrays have no duplicates.
 

Input: set1[] = {12, 34, 11, 9, 3}
       set2[] = {2, 1, 3, 5}
Output: Not Disjoint
3 is common in two sets.

Input: set1[] = {12, 34, 11, 9, 3}
       set2[] = {7, 2, 1, 5}
Output: Yes, Disjoint
There is no common element in two sets.

There are plenty of methods to solve this problem, it’s a good test to check how many solutions you can guess.
Method 1 (Simple) 
Iterate through every element of the first set and search it in another set, if any element is found, return false. If no element is found, return tree. The time complexity of this method is O(mn).
Following is the implementation of the above idea.
 

C++

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// A Simple C++ program to check if two sets are disjoint
#include<bits/stdc++.h>
using namespace std;
  
// Returns true if set1[] and set2[] are disjoint, else false
bool areDisjoint(int set1[], int set2[], int m, int n)
{
    // Take every element of set1[] and search it in set2
    for (int i=0; i<m; i++)
      for (int j=0; j<n; j++)
         if (set1[i] == set2[j])
            return false;
  
    // If no element of set1 is present in set2
    return true;
}
  
// Driver program to test above function
int main()
{
    int set1[] = {12, 34, 11, 9, 3};
    int set2[] = {7, 2, 1, 5};
    int m = sizeof(set1)/sizeof(set1[0]);
    int n = sizeof(set2)/sizeof(set2[0]);
    areDisjoint(set1, set2, m, n)? cout << "Yes" : cout << " No";
    return 0;
}

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Java

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// Java program to check if two sets are disjoint
  
public class disjoint1 
{
    // Returns true if set1[] and set2[] are 
    // disjoint, else false
    boolean aredisjoint(int set1[], int set2[]) 
    {
         // Take every element of set1[] and 
         // search it in set2
        for (int i = 0; i < set1.length; i++) 
        {
            for (int j = 0; j < set2.length; j++) 
            {
                if (set1[i] == set2[j])
                    return false;
            }
        }
        // If no element of set1 is present in set2
        return true;
    }
      
    // Driver program to test above function
    public static void main(String[] args) 
    {
        disjoint1 dis = new disjoint1();
        int set1[] = { 12, 34, 11, 9, 3 };
        int set2[] = { 7, 2, 1, 5 };
  
        boolean result = dis.aredisjoint(set1, set2);
        if (result)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
  
// This code is contributed by Rishabh Mahrsee

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Python

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# A Simple python 3 program to check
# if two sets are disjoint
  
# Returns true if set1[] and set2[] are disjoint, else false
def areDisjoint(set1, set2, m, n):
    # Take every element of set1[] and search it in set2
    for i in range(0, m):
        for j in range(0, n):
            if (set1[i] == set2[j]):
                return False
  
    # If no element of set1 is present in set2
    return True
  
  
# Driver program
set1 = [12, 34, 11, 9, 3]
set2 = [7, 2, 1, 5]
m = len(set1)
n = len(set2)
print("yes") if areDisjoint(set1, set2, m, n) else(" No")
  
# This code ia contributed by Smitha Dinesh Semwal

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C#

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// C# program to check if two 
// sets are disjoint 
using System;
  
class GFG
{
// Returns true if set1[] and set2[] 
// are disjoint, else false 
public virtual bool aredisjoint(int[] set1, 
                                int[] set2)
{
    // Take every element of set1[] 
    // and search it in set2 
    for (int i = 0; i < set1.Length; i++)
    {
        for (int j = 0; 
                 j < set2.Length; j++)
        {
            if (set1[i] == set2[j])
            {
                return false;
            }
        }
    }
      
    // If no element of set1 is 
    // present in set2 
    return true;
}
  
// Driver Code
public static void Main(string[] args)
{
    GFG dis = new GFG();
    int[] set1 = new int[] {12, 34, 11, 9, 3};
    int[] set2 = new int[] {7, 2, 1, 5};
  
    bool result = dis.aredisjoint(set1, set2);
    if (result)
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
  
// This code is contributed by Shrikant13

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PHP

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<?php
// A Simple PHP program to check 
// if two sets are disjoint
  
// Returns true if set1[] and set2[] 
// are disjoint, else false
function areDisjoint($set1, $set2, $m, $n)
{
    // Take every element of set1[]
    // and search it in set2
    for ($i = 0; $i < $m; $i++)
    for ($j = 0; $j < $n; $j++)
        if ($set1[$i] == $set2[$j])
            return false;
  
    // If no element of set1 is 
    // present in set2
    return true;
}
  
// Driver Code
$set1 = array(12, 34, 11, 9, 3);
$set2 = array(7, 2, 1, 5);
$m = sizeof($set1);
$n = sizeof($set2);
if(areDisjoint($set1, $set2,
               $m, $n) == true)
    echo "Yes";
else
    echo " No";
  
// This code is contributed
// by Akanksha Rai
?>

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Output : 

Yes

Method 2 (Use Sorting and Merging) 
1) Sort first and second sets. 
2) Use merge like the process to compare elements.
Following is the implementation of the above idea.
 

C++

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// A Simple C++ program to check if two sets are disjoint
#include<bits/stdc++.h>
using namespace std;
  
// Returns true if set1[] and set2[] are disjoint, else false
bool areDisjoint(int set1[], int set2[], int m, int n)
{
    // Sort the given two sets
    sort(set1, set1+m);
    sort(set2, set2+n);
  
    // Check for same elements using merge like process
    int i = 0, j = 0;
    while (i < m && j < n)
    {
        if (set1[i] < set2[j])
            i++;
        else if (set2[j] < set1[i])
            j++;
        else /* if set1[i] == set2[j] */
            return false;
    }
  
    return true;
}
  
// Driver program to test above function
int main()
{
    int set1[] = {12, 34, 11, 9, 3};
    int set2[] = {7, 2, 1, 5};
    int m = sizeof(set1)/sizeof(set1[0]);
    int n = sizeof(set2)/sizeof(set2[0]);
    areDisjoint(set1, set2, m, n)? cout << "Yes" : cout << " No";
    return 0;
}

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Java

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// Java program to check if two sets are disjoint
  
import java.util.Arrays;
  
public class disjoint1 
{
    // Returns true if set1[] and set2[] are 
    // disjoint, else false
    boolean aredisjoint(int set1[], int set2[]) 
    {
        int i=0,j=0;
          
        // Sort the given two sets
        Arrays.sort(set1);
        Arrays.sort(set2);
          
        // Check for same elements using 
        // merge like process
        while(i<set1.length && j<set2.length)
        {
            if(set1[i]<set2[j])
                i++;
            else if(set1[i]>set2[j])
                j++;
            else 
                return false;
              
        }
        return true;
    }
  
    // Driver program to test above function
    public static void main(String[] args) 
    {
        disjoint1 dis = new disjoint1();
        int set1[] = { 12, 34, 11, 9, 3 };
        int set2[] = { 7, 2, 1, 5 };
  
        boolean result = dis.aredisjoint(set1, set2);
        if (result)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}
  
// This code is contributed by Rishabh Mahrsee

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Python

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# A Simple Python 3 program to check
# if two sets are disjoint
  
# Returns true if set1[] and set2[]
# are disjoint, else false
def areDisjoint(set1, set2, m, n):
    # Sort the given two sets
    set1.sort()
    set2.sort()
  
    # Check for same elements  
    # using merge like process
    i = 0; j = 0
    while (i < m and j < n):
          
        if (set1[i] < set2[j]):
            i += 1
        elif (set2[j] < set1[i]):
            j += 1
        else: # if set1[i] == set2[j] 
            return False
    return True
  
  
# Driver Code
set1 = [12, 34, 11, 9, 3]
set2 = [7, 2, 1, 5]
m = len(set1)
n = len(set2)
  
print("Yes") if areDisjoint(set1, set2, m, n) else print("No")
  
# This code is contributed by Smitha Dinesh Semwal

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C#

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// C# program to check if two sets are disjoint
using System;
  
public class disjoint1 
{
    // Returns true if set1[] and set2[] are 
    // disjoint, else false
    Boolean aredisjoint(int []set1, int []set2) 
    {
        int i = 0, j = 0;
          
        // Sort the given two sets
        Array.Sort(set1);
        Array.Sort(set2);
          
        // Check for same elements using 
        // merge like process
        while(i < set1.Length && j < set2.Length)
        {
            if(set1[i] < set2[j])
                i++;
            else if(set1[i] > set2[j])
                j++;
            else
                return false;
              
        }
        return true;
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        disjoint1 dis = new disjoint1();
        int []set1 = { 12, 34, 11, 9, 3 };
        int []set2 = { 7, 2, 1, 5 };
  
        Boolean result = dis.aredisjoint(set1, set2);
        if (result)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}
  
// This code contributed by Rajput-Ji

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Output : 



Yes

Time complexity of above solution is O(mLogm + nLogn). 
The above solution first sorts both sets then takes O(m+n) time to find the intersection. If we are given that the input sets are sorted, then this method is best among all.
Method 3 (Use Sorting and Binary Search) 
This is similar to method 1. Instead of a linear search, we use Binary Search
1) Sort first set. 
2) Iterate through every element of the second set, and use binary search to search every element in the first set. If an element is found return it.
The time complexity of this method is O(mLogm + nLogm)
Method 4 (Use Binary Search Tree) 
1) Create a self-balancing binary search tree (Red Black, AVL, Splay, etc) of all elements in the first set. 
2) Iterate through all elements of the second set and search every element in the above constructed Binary Search Tree. If the element is found, return false. 
3) If all elements are absent, return true.
The time complexity of this method is O(mLogm + nLogm). 
Method 5 (Use Hashing) 
1) Create an empty hash table. 
2) Iterate through the first set and store every element in the hash table. 
3) Iterate through the second set and check if any element is present in the hash table. If present, then returns false, else ignore the element. 
4) If all elements of the second set are not present in the hash table, return true.
The following are the implementation of this method. 
 

C/C++

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/* C++ program to check if two sets are distinct or not */
#include<bits/stdc++.h>
using namespace std;
  
// This function prints all distinct elements
bool areDisjoint(int set1[], int set2[], int n1, int n2)
{
    // Creates an empty hashset
    set<int> myset;
  
    // Traverse the first set and store its elements in hash
    for (int i = 0; i < n1; i++)
        myset.insert(set1[i]);
  
    // Traverse the second set and check if any element of it
    // is already in hash or not.
    for (int i = 0; i < n2; i++)
        if (myset.find(set2[i]) != myset.end())
            return false;
  
    return true;
}
  
// Driver method to test above method
int main()
{
    int set1[] = {10, 5, 3, 4, 6};
    int set2[] = {8, 7, 9, 3};
  
    int n1 = sizeof(set1) / sizeof(set1[0]);
    int n2 = sizeof(set2) / sizeof(set2[0]);
    if (areDisjoint(set1, set2, n1, n2))
        cout << "Yes";
    else
        cout << "No";
}
//This article is contributed by Chhavi

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Java

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/* Java program to check if two sets are distinct or not */
import java.util.*;
  
class Main
{
    // This function prints all distinct elements
    static boolean areDisjoint(int set1[], int set2[])
    {
        // Creates an empty hashset
        HashSet<Integer> set = new HashSet<>();
  
        // Traverse the first set and store its elements in hash
        for (int i=0; i<set1.length; i++)
            set.add(set1[i]);
  
        // Traverse the second set and check if any element of it
        // is already in hash or not.
        for (int i=0; i<set2.length; i++)
            if (set.contains(set2[i]))
                return false;
  
        return true;
    }
  
    // Driver method to test above method
    public static void main (String[] args)
    {
        int set1[] = {10, 5, 3, 4, 6};
        int set2[] = {8, 7, 9, 3};
        if (areDisjoint(set1, set2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

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Python3

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# Python3 program to 
# check if two sets are 
# distinct or not 
# This function prints 
# all distinct elements
def areDisjoint(set1, set2, 
                n1, n2):
    
  # Creates an empty hashset
  myset = set([])
    
  # Traverse the first set 
  # and store its elements in hash
  for i in range (n1):
    myset.add(set1[i])
      
  # Traverse the second set 
  # and check if any element of it
  # is already in hash or not.
  for i in range (n2):
    if (set2[i] in myset):
      return False
  return True
  
# Driver method to test above method
if __name__ == "__main__":
    
  set1 = [10, 5, 3, 4, 6]
  set2 = [8, 7, 9, 3]
  
  n1 = len(set1)
  n2 = len(set2)
    
  if (areDisjoint(set1, set2,
                  n1, n2)):
    print ("Yes")
  else:
    print("No")
  
# This code is contributed by Chitranayal

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C#

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using System;
using System.Collections.Generic;
  
/* C# program to check if two sets are distinct or not */
  
public class GFG
{
    // This function prints all distinct elements 
    public static bool areDisjoint(int[] set1, int[] set2)
    {
        // Creates an empty hashset 
        HashSet<int> set = new HashSet<int>();
  
        // Traverse the first set and store its elements in hash 
        for (int i = 0; i < set1.Length; i++)
        {
            set.Add(set1[i]);
        }
  
        // Traverse the second set and check if any element of it 
        // is already in hash or not. 
        for (int i = 0; i < set2.Length; i++)
        {
            if (set.Contains(set2[i]))
            {
                return false;
            }
        }
  
        return true;
    }
  
    // Driver method to test above method 
    public static void Main(string[] args)
    {
        int[] set1 = new int[] {10, 5, 3, 4, 6};
        int[] set2 = new int[] {8, 7, 9, 3};
        if (areDisjoint(set1, set2))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
//This code is contributed by Shrikant13

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Output: 

No

Time complexity of the above implementation is O(m+n) under the assumption that hash set operations like add() and contains() work in O(1) time.
This article is contributed by Rajeev. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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