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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 first set and search it in other set, if any element is found, return false. If no element is found, return tree. Time complexity of this method is O(mn).

Following is implementation of above idea.

C++

 // A Simple C++ program to check if two sets are disjoint #include 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

Java

 // 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

Python

 # 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#

 // 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

PHP



Output :

Yes

Method 2 (Use Sorting and Merging)
1) Sort first and second sets.
2) Use merge like process to compare elements.

Following is implementation of above idea.

C++

 // A Simple C++ program to check if two sets are disjoint #include #include 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);     int n = sizeof(set2)/sizeof(set2);     areDisjoint(set1, set2, m, n)? cout << "Yes" : cout << " No";     return 0; }

Java

 // 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(iset2[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

Python

 # 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

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 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 linear search, we use Binary Search.
1) Sort first set.
2) Iterate through every element of second set, and use binary search to search every element in first set. If element is found return it.

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 first set.
2) Iterate through all elements of second set and search every element in the above constructed Binary Search Tree. If element is found, return false.
3) If all elements are absent, return true.

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 hash table.
3) Iterate through second set and check if any element is present in hash table. If present, then return false, else ignore the element.
4) If all elements of second set are not present in hash table, return true.

Following are the implementation of this method.

C/C++

 #include using namespace std;    /* C++ program to check if two sets are distinct or not */ // This function prints all distinct elements bool areDisjoint(int set1[], int set2[], int n1, int n2) {     // Creates an empty hashset     set 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);     int n2 = sizeof(set2) / sizeof(set2);     if (areDisjoint(set1, set2, n1, n2))         cout << "Yes";     else         cout << "No"; } //This article is contributed by Chhavi

Java

 /* 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 set = new HashSet<>();            // Traverse the first set and store its elements in hash         for (int i=0; i

C#

 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 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 = 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

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.