# 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 `

## 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 `

## C#

 `// 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 `

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<``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); ` `    ``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<``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 `

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.