# Check if two unsorted arrays (with duplicates allowed) have same elements

Given two unsorted arrays, check whether both arrays have same set of elements or not.
Examples:

```Input : A = {2, 5, 6, 8, 10, 2, 2}
B = {2, 5, 5, 6, 8, 5, 6}
Output : No

Input : A = {2, 5, 6, 8, 2, 10, 2}
B = {2, 5, 6, 8, 2, 10, 2}
Output : Yes

Input : A = {2, 5, 8, 6, 10, 2, 2}
B = {2, 5, 6, 8, 2, 10, 2}
Output : Yes
```

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

Method 1 (Simple):

A simple solution to this problem is to check if each element of A is present in B. But this approach will lead to wrong answer in case of multiple instances of an element is present in B. To overcome this issue, we mark visited instances of B[] using an auxiliary array visited[].

## C++

 `#include ` `using` `namespace` `std; ` `// Function to check if both arrays are same ` `bool` `areSameSet(vector<``int``> A, vector<``int``> B) ` `{ ` `    ``int` `n = A.size(); ` `    ``if` `(B.size() != n) ` `        ``return` `false``; ` ` `  `    ``// visited array is used to handle duplicates  ` `    ``vector<``bool``> visited(n, ``false``); ` ` `  `    ``// each element of A is matched ` `    ``// against each element of B ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``int` `j = 0; ` `        ``for` `(j = 0; j < n; j++)  ` `        ``{  ` `            ``if` `(A[i] == B[j] && visited[j] == ``false``)  ` `            ``{ ` `                ``visited[j] = ``true``; ` `                ``break``;             ` `            ``}    ` `        ``} ` ` `  `        ``// If we could not find A[i] in B[] ` `        ``if` `(j == n) ` `            ``return` `false``;    ` `        `  `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``int``> A, B; ` `    ``A.push_back(2); ` `    ``A.push_back(5); ` `    ``A.push_back(10); ` `    ``A.push_back(6); ` `    ``A.push_back(8); ` `    ``A.push_back(2); ` `    ``A.push_back(2); ` ` `  `    ``B.push_back(2); ` `    ``B.push_back(5); ` `    ``B.push_back(6); ` `    ``B.push_back(8); ` `    ``B.push_back(10); ` `    ``B.push_back(2); ` `    ``B.push_back(2); ` ` `  `    ``areSameSet(A, B)? cout << ``"Yes"` `: cout << ``"No"``; ` `} `

## Java

 `// Java implementation of the above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to check if both arrays are same ` `    ``static` `boolean` `areSameSet(Vector A, Vector B)  ` `    ``{ ` `        ``int` `n = A.size(); ` `        ``if` `(B.size() != n)  ` `        ``{ ` `            ``return` `false``; ` `        ``} ` ` `  `        ``// visited array is used to handle duplicates  ` `        ``Vector visited = ``new` `Vector(); ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``visited.add(i, Boolean.FALSE); ` `        ``} ` `         `  `        ``// each element of A is matched ` `        ``// against each element of B ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` ` `  `            ``int` `j = ``0``; ` `            ``for` `(j = ``0``; j < n; j++) ` `            ``{ ` `                ``if` `(A.get(i) == B.get(j) && visited.get(j) == ``false``)  ` `                ``{ ` `                    ``visited.add(j, Boolean.TRUE); ` `                    ``break``; ` `                ``} ` `            ``} ` ` `  `            ``// If we could not find A[i] in B[] ` `            ``if` `(j == n) ` `            ``{ ` `                ``return` `false``; ` `            ``} ` ` `  `        ``} ` `        ``return` `true``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``Vector A = ``new` `Vector<>(); ` `        ``Vector B = ``new` `Vector<>(); ` `        ``A.add(``2``); ` `        ``A.add(``5``); ` `        ``A.add(``10``); ` `        ``A.add(``6``); ` `        ``A.add(``8``); ` `        ``A.add(``2``); ` `        ``A.add(``2``); ` ` `  `        ``B.add(``2``); ` `        ``B.add(``5``); ` `        ``B.add(``6``); ` `        ``B.add(``8``); ` `        ``B.add(``10``); ` `        ``B.add(``2``); ` `        ``B.add(``2``); ` ` `  `        ``if` `(areSameSet(A, B))  ` `        ``{ ` `            ``System.out.println(``"Yes"``); ` `        ``}  ` `        ``else` `        ``{ ` `            ``System.out.println(``"No"``); ` `        ``} ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Function to check if both arrays are same ` `def` `areSameSet(A, B): ` ` `  `    ``n ``=` `len``(A) ` `    ``if` `(``len``(B) !``=` `n): ` `        ``return` `False` ` `  `    ``# visited array is used to handle duplicates ` `    ``visited ``=` `[``False` `for` `i ``in` `range``(n)] ` ` `  `    ``# each element of A is matched ` `    ``# against each element of B ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``j ``=` `0` `        ``for` `j ``in` `range``(n): ` `            ``if` `(A[i] ``=``=` `B[j] ``and`  `                ``visited[j] ``=``=` `False``): ` `                ``visited[j] ``=` `True` `                ``break` ` `  `        ``# If we could not find A[i] in B[] ` `        ``if` `(j ``=``=` `n): ` `            ``return` `False` ` `  `    ``return` `True` ` `  `# Driver code ` `A ``=` `[] ` `B ``=` `[] ` `A.append(``2``) ` `A.append(``5``) ` `A.append(``10``) ` `A.append(``6``) ` `A.append(``8``) ` `A.append(``2``) ` `A.append(``2``) ` ` `  `B.append(``2``) ` `B.append(``5``) ` `B.append(``6``) ` `B.append(``8``) ` `B.append(``10``) ` `B.append(``2``) ` `B.append(``2``) ` ` `  `if``(areSameSet(A, B)): ` `    ``print``(``"Yes"``) ` `else``: ` `    ``print``(``"No"``) ` `     `  `# This code is contributed  ` `# by mohit kumar `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``// Function to check if both arrays are same  ` `    ``static` `Boolean areSameSet(List<``int``> A, List<``int``> B)  ` `    ``{  ` `        ``int` `n = A.Count;  ` `        ``if` `(B.Count != n)  ` `        ``{  ` `            ``return` `false``;  ` `        ``}  ` ` `  `        ``// visited array is used to handle duplicates  ` `        ``List visited = ``new` `List();  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `            ``visited.Insert(i, ``false``);  ` `        ``}  ` `         `  `        ``// each element of A is matched  ` `        ``// against each element of B  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` ` `  `            ``int` `j = 0;  ` `            ``for` `(j = 0; j < n; j++)  ` `            ``{  ` `                ``if` `(A[i] == B[j] && visited[j] == ``false``)  ` `                ``{  ` `                    ``visited.Insert(j, ``true``);  ` `                    ``break``;  ` `                ``}  ` `            ``}  ` ` `  `            ``// If we could not find A[i] in B[]  ` `            ``if` `(j == n)  ` `            ``{  ` `                ``return` `false``;  ` `            ``}  ` ` `  `        ``}  ` `        ``return` `true``;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{  ` `        ``List<``int``> A = ``new` `List<``int``>();  ` `        ``List<``int``> B = ``new` `List<``int``>();  ` `        ``A.Add(2);  ` `        ``A.Add(5);  ` `        ``A.Add(10);  ` `        ``A.Add(6);  ` `        ``A.Add(8);  ` `        ``A.Add(2);  ` `        ``A.Add(2);  ` ` `  `        ``B.Add(2);  ` `        ``B.Add(5);  ` `        ``B.Add(6);  ` `        ``B.Add(8);  ` `        ``B.Add(10);  ` `        ``B.Add(2);  ` `        ``B.Add(2);  ` ` `  `        ``if` `(areSameSet(A, B))  ` `        ``{  ` `            ``Console.WriteLine(``"Yes"``);  ` `        ``}  ` `        ``else` `        ``{  ` `            ``Console.WriteLine(``"No"``);  ` `        ``}  ` `    ``}  ` `}  ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```Yes
```

The time complexity of above solution in O(n^2).

Method 2 (Sorting) :

Sort both the arrays and compare corresponding elements of each array.

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// Function to check if both arrays are same ` `bool` `areSameSet(vector<``int``> A, vector<``int``> B) ` `{ ` `    ``int` `n = A.size(); ` `    ``if` `(B.size() != n) ` `        ``return` `false``; ` ` `  `    ``sort(A.begin(), A.end()); ` `    ``sort(B.begin(), B.end()); ` ` `  `    ``// Compare corresponding elements ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``if` `(A[i] != B[i])  ` `            ``return` `false``;         ` `     `  `    ``return` `true``; ` `} ` ` `  `int` `main() ` `{ ` `    ``vector<``int``> A, B; ` `    ``A.push_back(2); ` `    ``A.push_back(5); ` `    ``A.push_back(10); ` `    ``A.push_back(6); ` `    ``A.push_back(8); ` `    ``A.push_back(2); ` `    ``A.push_back(2); ` ` `  `    ``B.push_back(2); ` `    ``B.push_back(5); ` `    ``B.push_back(6); ` `    ``B.push_back(8); ` `    ``B.push_back(10); ` `    ``B.push_back(2); ` `    ``B.push_back(2); ` ` `  `    ``areSameSet(A, B)? cout << ``"Yes"` `: cout << ``"No"``; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `    ``// Function to check if both arrays are same ` `    ``static` `boolean` `areSameSet(Vector A,  ` `                                ``Vector B)  ` `    ``{ ` `        ``int` `n = A.size(); ` `        ``if` `(B.size() != n)  ` `        ``{ ` `            ``return` `false``; ` `        ``} ` ` `  `        ``Collections.sort(A); ` `        ``Collections.sort(B); ` ` `  `        ``// Compare corresponding elements ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``if` `(A.get(i) != B.get(i))  ` `            ``{ ` `                ``return` `false``; ` `            ``} ` `        ``} ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``Vector A = ``new` `Vector<>(); ` `        ``Vector B = ``new` `Vector<>(); ` `        ``A.add(``2``); ` `        ``A.add(``5``); ` `        ``A.add(``10``); ` `        ``A.add(``6``); ` `        ``A.add(``8``); ` `        ``A.add(``2``); ` `        ``A.add(``2``); ` ` `  `        ``B.add(``2``); ` `        ``B.add(``5``); ` `        ``B.add(``6``); ` `        ``B.add(``8``); ` `        ``B.add(``10``); ` `        ``B.add(``2``); ` `        ``B.add(``2``); ` ` `  `        ``if` `(areSameSet(A, B))  ` `        ``{ ` `            ``System.out.println(``"Yes"``); ` `        ``}  ` `        ``else` `        ``{ ` `            ``System.out.println(``"No"``); ` `        ``} ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG  ` `{ ` `    ``// Function to check if both arrays are same ` `    ``static` `Boolean areSameSet(List<``int``> A,  ` `                                ``List<``int``> B)  ` `    ``{ ` `        ``int` `n = A.Count; ` `        ``if` `(B.Count!= n)  ` `        ``{ ` `            ``return` `false``; ` `        ``} ` ` `  `        ``A.Sort(); ` `        ``B.Sort(); ` ` `  `        ``// Compare corresponding elements ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `(A[i] != B[i])  ` `            ``{ ` `                ``return` `false``; ` `            ``} ` `        ``} ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``List<``int``> A = ``new` `List<``int``>(); ` `        ``List<``int``> B = ``new` `List<``int``>(); ` `        ``A.Add(2); ` `        ``A.Add(5); ` `        ``A.Add(10); ` `        ``A.Add(6); ` `        ``A.Add(8); ` `        ``A.Add(2); ` `        ``A.Add(2); ` ` `  `        ``B.Add(2); ` `        ``B.Add(5); ` `        ``B.Add(6); ` `        ``B.Add(8); ` `        ``B.Add(10); ` `        ``B.Add(2); ` `        ``B.Add(2); ` ` `  `        ``if` `(areSameSet(A, B))  ` `        ``{ ` `            ``Console.WriteLine(``"Yes"``); ` `        ``}  ` `        ``else` `        ``{ ` `            ``Console.WriteLine(``"No"``); ` `        ``} ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

Output:

`Yes`

The time Complexity of above solution in O(n*log(n)).

Method 3 (Hashing):

We can decrease the time complexity of the above problem by using a Hash table. First, we iterate through A and mark the number of instances of each element of A in a Hash Table. Then we iterate through B and decrease the corresponding value in the hash table. If in the end if all the entries of the hash table are zero, the answer will be “Yes” else “No”.

## C++

 `#include ` `using` `namespace` `std; ` ` `  `bool` `areSameSet(vector<``int``> A, vector<``int``> B) ` `{ ` `    ``int` `n = A.size(); ` `    ``if` `(B.size() != n) ` `        ``return` `false``; ` ` `  `    ``// Create a hash table to ` `    ``// number of instances ` `    ``unordered_map<``int``> m; ` ` `  `    ``// for each element of A ` `    ``// increase it's instance by 1. ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``m[A[i]]++; ` `     `  `    ``// for each element of B ` `    ``// decrease it's instance by 1. ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``m[B[i]]--; ` `     `  `    ``// Iterate through map and check if  ` `    ``// any entry is non-zero ` `    ``for` `(``auto` `i : m)  ` `        ``if` `(i.second != 0) ` `            ``return` `false``; ` `     `  `    ``return` `true``; ` `} ` ` `  `int` `main() ` `{ ` `    ``vector<``int``> A, B; ` `    ``A.push_back(2); ` `    ``A.push_back(5); ` `    ``A.push_back(10); ` `    ``A.push_back(6); ` `    ``A.push_back(8); ` `    ``A.push_back(2); ` `    ``A.push_back(2); ` ` `  `    ``B.push_back(2); ` `    ``B.push_back(5); ` `    ``B.push_back(6); ` `    ``B.push_back(8); ` `    ``B.push_back(10); ` `    ``B.push_back(2); ` `    ``B.push_back(2); ` ` `  `    ``areSameSet(A, B)? cout << ``"Yes"` `: cout << ``"No"``; ` `} `

## Java

 `// Java program to implement Naive approach  ` `// to remove duplicates ` `import` `java.util.HashMap; ` ` `  `class` `GFG  ` `{ ` `static` `boolean` `areSameSet(``int``[] A, ``int``[] B)  ` `{ ` `    ``int` `n = A.length; ` ` `  `    ``if` `(B.length != n) ` `        ``return` `false``; ` ` `  `    ``// Create a hash table to ` `    ``// number of instances ` `    ``HashMap m = ``new` `HashMap<>(); ` ` `  `    ``// for each element of A ` `    ``// increase it's instance by 1. ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``m.put(A[i], m.get(A[i]) == ``null` `? ``1` `:  ` `                    ``m.get(A[i]) + ``1``); ` ` `  `    ``// for each element of B ` `    ``// decrease it's instance by 1. ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``m.put(B[i], m.get(B[i]) - ``1``); ` ` `  `    ``// Iterate through map and check if ` `    ``// any entry is non-zero ` `    ``for` `(HashMap.Entry entry : m.entrySet()) ` `        ``if` `(entry.getValue() != ``0``) ` `            ``return` `false``; ` `    ``return` `true``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int``[] A = { ``2``, ``5``, ``10``, ``6``, ``8``, ``2``, ``2` `}; ` `    ``int``[] B = { ``2``, ``5``, ``6``, ``8``, ``10``, ``2``, ``2` `}; ` ` `  `    ``if` `(areSameSet(A, B)) ` `        ``System.out.println(``"Yes"``); ` `    ``else` `        ``System.out.println(``"No"``); ` `} ` `} ` ` `  `// This code is contributed by ` `// sanjeev2552 `

Output:

```Yes
```

The time complexity of above method is O(n).

This article is contributed by Raghav Sharma. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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