Open in App
Not now

# How to check if two given sets are disjoint?

• Difficulty Level : Easy
• Last Updated : 25 Mar, 2023

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 true. The time complexity of this method is O(mn).

Following is the implementation of the 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

 ``

## Javascript

 ``

Output

`Yes`

Time Complexity: O(m*n)
Auxiliary Space: O(1),As constant extra space is used.

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++

 `// 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)` `{` `    ``// 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;` `}`

## 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`

## Python3

 `# 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`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(m*log m + n*log n), The above solution first sorts both sets and 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.
Auxiliary Space: O(1), As constant extra space is used.

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(m*log m + n*log m).

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 is the implementation of this method.

## C++

 `/* C++ program to check if two sets are distinct or not */` `#include` `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`

## 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

## Python3

 `# 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`

## 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`

## Javascript

 ``

Output

`No`

Time Complexity: O(m+n) under the assumption that hash set operations like add() and contains() work in O(1) time.
Auxiliary Space: O(n), The extra space is used to store the elements in the set.

Method 6: Using inbuilt function

## C++

 `#include ` `#include ` `using` `namespace` `std;`   `bool` `areDisjoint(``int` `arr1[], ``int` `n1, ``int` `arr2[], ``int` `n2)` `{` `    ``unordered_set<``int``> set1(arr1, arr1 + n1);` `    ``unordered_set<``int``> set2(arr2, arr2 + n2);` `    ``unordered_set<``int``> result;` `    ``for` `(``int` `x : set1) {` `        ``if` `(set2.find(x) != set2.end()) {` `            ``result.insert(x);` `        ``}` `    ``}` `    ``if` `(result.size() != 0) {` `        ``return` `false``;` `    ``}` `    ``else` `{` `        ``return` `true``;` `    ``}` `}`   `int` `main()` `{` `    ``int` `arr1[] = { 10, 5, 3, 4, 6 };` `    ``int` `n1 = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` `    ``int` `arr2[] = { 8, 7, 9, 3 };` `    ``int` `n2 = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` `    ``if` `(areDisjoint(arr1, n1, arr2, n2)) {` `        ``cout << ``"YES"` `<< endl;` `    ``}` `    ``else` `{` `        ``cout << ``"NO"` `<< endl;` `    ``}` `    ``return` `0;` `}`

## Java

 `// Import necessary Java libraries` `import` `java.util.Arrays;` `import` `java.util.HashSet;` `import` `java.util.Set;`   `public` `class` `DisjointArrays {` `    ``public` `static` `boolean` `areDisjoint(``int``[] arr1, ``int` `n1,` `                                      ``int``[] arr2, ``int` `n2)` `    ``{`   `        ``// Create HashSet objects for arr1 and arr2` `        ``Set set1 = ``new` `HashSet();` `        ``for` `(``int` `i = ``0``; i < n1; i++) {` `            ``set1.add(arr1[i]);` `        ``}`   `        ``Set set2 = ``new` `HashSet();` `        ``for` `(``int` `i = ``0``; i < n2; i++) {` `            ``set2.add(arr2[i]);` `        ``}`   `        ``// Create a HashSet to store common elements` `        ``Set result = ``new` `HashSet();` `        ``for` `(``int` `x : set1) {` `            ``if` `(set2.contains(x)) {` `                ``result.add(x);` `            ``}` `        ``}`   `        ``// If result set is not empty, then arrays are not` `        ``// disjoint` `        ``if` `(!result.isEmpty()) {` `            ``return` `false``;` `        ``}` `        ``else` `{` `            ``return` `true``;` `        ``}` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int``[] arr1 = { ``10``, ``5``, ``3``, ``4``, ``6` `};` `        ``int` `n1 = arr1.length;`   `        ``int``[] arr2 = { ``8``, ``7``, ``9``, ``3` `};` `        ``int` `n2 = arr2.length;`   `        ``// Check if arrays are disjoint` `        ``if` `(areDisjoint(arr1, n1, arr2, n2)) {` `            ``System.out.println(``"YES"``);` `        ``}` `        ``else` `{` `            ``System.out.println(``"NO"``);` `        ``}` `    ``}` `}`

## Python3

 `def` `areDisjoint(arr1, arr2):` `  ``set1 ``=` `set``(arr1)` `  ``set2 ``=` `set``(arr2)` `  ``result ``=` `set1.intersection(set2)` `  ``if` `len``(result) !``=` `0``:` `    ``return` `False` `  ``else``:` `    ``return` `True` `arr1 ``=` `[``10``, ``5``, ``3``, ``4``, ``6``]` `arr2 ``=` `[``8``, ``7``, ``9``, ``3``]` `if` `areDisjoint(arr1, arr2):` `  ``print``(``"YES"``)` `else``:` `  ``print``(``"NO"``)`   `# This code is contributed by Prince Kumar`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `class` `Program {` `    ``// Method to check if two arrays are disjoint` `    ``static` `bool` `AreDisjoint(``int``[] arr1, ``int``[] arr2) {` `        ``// Create hash sets from the input arrays` `        ``HashSet<``int``> set1 = ``new` `HashSet<``int``>(arr1);` `        ``HashSet<``int``> set2 = ``new` `HashSet<``int``>(arr2);` `        ``// Initialize an empty hash set to store common elements` `        ``HashSet<``int``> result = ``new` `HashSet<``int``>();` `        ``// Iterate through elements of set1` `        ``foreach` `(``int` `x ``in` `set1) {` `            ``// If x is present in set2, add it to result` `            ``if` `(set2.Contains(x)) {` `                ``result.Add(x);` `            ``}` `        ``}` `        ``// If result is empty, the arrays are disjoint` `        ``return` `result.Count == 0;` `    ``}`   `    ``static` `void` `Main(``string``[] args) {` `        ``// Define example arrays` `        ``int``[] arr1 = { 10, 5, 3, 4, 6 };` `        ``int``[] arr2 = { 8, 7, 9, 3 };` `        ``// Check if the arrays are disjoint` `        ``if` `(AreDisjoint(arr1, arr2)) {` `            ``Console.WriteLine(``"YES"``);` `        ``}` `        ``else` `{` `            ``Console.WriteLine(``"NO"``);` `        ``}` `    ``}` `}`

## Javascript

 `function` `areDisjoint(arr1, n1, arr2, n2) {` `    ``// Create sets from the two input arrays` `    ``let set1 = ``new` `Set(arr1);` `    ``let set2 = ``new` `Set(arr2);`   `    ``// Check for common elements` `    ``for` `(let x of set1) {` `        ``if` `(set2.has(x)) {` `            ``return` `false``;` `        ``}` `    ``}` `    ``return` `true``;` `}`   `let arr1 = [10, 5, 3, 4, 6];` `let n1 = arr1.length;` `let arr2 = [8, 7, 9, 3];` `let n2 = arr2.length;`   `if` `(areDisjoint(arr1,n1,arr2,n2)) {` `    ``console.log(``"YES"``);` `} ``else` `{` `    ``console.log(``"NO"``);` `}`

Output

`NO`

Time Complexity: O(min(n, m)), where n = length of arr1, m = length of arr2

Auxiliary Space: O(max(n, m))

My Personal Notes arrow_drop_up
Related Articles