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# Check if two arrays are permutations of each other

Given two unsorted arrays of the same size, write a function that returns true if two arrays are permutations of each other, otherwise false.

Examples:

```Input: arr1[] = {2, 1, 3, 5, 4, 3, 2}
arr2[] = {3, 2, 2, 4, 5, 3, 1}
Output: Yes

Input: arr1[] = {2, 1, 3, 5,}
arr2[] = {3, 2, 2, 4}
Output: No```

We strongly recommend you to minimize your browser and try this yourself first.
A Simple Solution is to sort both arrays and compare sorted arrays. The time complexity of this solution is O(nLogn)
A Better Solution is to use Hashing

1. Create a Hash Map for all the elements of arr1[] such that array elements are keys and their counts are values.
2. Traverse arr2[] and search for each element of arr2[] in the Hash Map. If an element is found then decrement its count in the hash map. If not found, then return false.
3. If all elements are found then return true.

Below is the implementation of this approach.

## C++

 `// A C++ program to find one array is permutation of other array``#include ``using` `namespace` `std;` `// Returns true if arr1[] and arr2[] are permutations of each other``bool` `arePermutations(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m)``{``    ` `    ``// Arrays cannot be permutations of one another unless they are``    ``// of the same length``    ``if``(n != m)``    ``{``        ``return` `false``;``    ``}``  ` `    ``// Creates an empty hashMap hM``    ``unordered_map<``int``, ``int``> hm;``    ` `    ``// Traverse through the first array and add elements to hash map``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``int` `x = arr1[i];``        ``hm[x]++;``    ``}``    ` `    ``// Traverse through second array and check if every element is``    ``// present in hash map``    ``for` `(``int` `i = 0; i < m; i++)``    ``{``        ``int` `x = arr2[i];``      ` `        ``// If element is not present in hash map or element``        ``// is not present less number of times``        ``if``(hm[x] == 0)``        ``{``            ``return` `false``;``        ``}``        ``hm[x]--;``    ``}``    ``return` `true``;``}` `// Driver function to test above function``int` `main() {``    ``int` `arr1[] = {2, 1, 3, 5, 4, 3, 2};``    ``int` `arr2[] = {3, 2, 2, 4, 5, 3, 1};``    ``int` `n = ``sizeof``(arr1)/``sizeof``(arr1[0]);``    ``int` `m = ``sizeof``(arr2)/``sizeof``(arr2[0]);``    ``if` `(arePermutations(arr1, arr2, n, m))``        ``cout << ``"Arrays are permutations of each other"` `<< endl;``    ``else``        ``cout << ``"Arrays are NOT permutations of each other"` `<< endl;``    ` `    ``return` `0;``}` `// This code is contributed by avanitrachhadiya2155`

## Java

 `// A Java program to find one array is permutation of other array``import` `java.util.HashMap;` `class` `Permutations``{``    ``// Returns true if arr1[] and arr2[] are permutations of each other``    ``static` `Boolean arePermutations(``int` `arr1[], ``int` `arr2[])``    ``{``        ``// Arrays cannot be permutations of one another unless they are``        ``// of the same length``        ``if` `(arr1.length != arr2.length)``            ``return` `false``;``      ` `        ``// Creates an empty hashMap hM``        ``HashMap hM = ``new` `HashMap();` `        ``// Traverse through the first array and add elements to hash map``        ``for` `(``int` `i = ``0``; i < arr1.length; i++)``        ``{``            ``int` `x = arr1[i];``            ``if` `(hM.get(x) == ``null``)``                ``hM.put(x, ``1``);``            ``else``            ``{``                ``int` `k = hM.get(x);``                ``hM.put(x, k+``1``);``            ``}``        ``}` `        ``// Traverse through second array and check if every element is``        ``// present in hash map``        ``for` `(``int` `i = ``0``; i < arr2.length; i++)``        ``{``            ``int` `x = arr2[i];` `            ``// If element is not present in hash map or element``            ``// is not present less number of times``            ``if` `(hM.get(x) == ``null` `|| hM.get(x) == ``0``)``                ``return` `false``;` `            ``int` `k = hM.get(x);``            ``hM.put(x, k-``1``);``        ``}``        ``return` `true``;``    ``}` `    ``// Driver function to test above function``    ``public` `static` `void` `main(String arg[])``    ``{``        ``int` `arr1[] = {``2``, ``1``, ``3``, ``5``, ``4``, ``3``, ``2``};``        ``int` `arr2[] = {``3``, ``2``, ``2``, ``4``, ``5``, ``3``, ``1``};``        ``if` `(arePermutations(arr1, arr2))``            ``System.out.println(``"Arrays are permutations of each other"``);``        ``else``            ``System.out.println(``"Arrays are NOT permutations of each other"``);``    ``}``}`

## Python3

 `# Python3 program to find one``# array is permutation of other array``from` `collections ``import` `defaultdict` `# Returns true if arr1[] and``# arr2[] are permutations of``# each other``def` `arePermutations(arr1, arr2):``    ``# Arrays cannot be permutations of one another``    ``# unless they are of the same length``    ``if` `(``len``(arr1) !``=` `len``(arr2)):``        ``return` `False``      ` `    ``# Creates an empty hashMap hM``    ``hM ``=` `defaultdict (``int``)` `    ``# Traverse through the first``    ``# array and add elements to``    ``# hash map``    ``for` `i ``in` `range` `(``len``(arr1)):``        ` `        ``x ``=` `arr1[i]``        ``hM[x] ``+``=` `1``        ` `    ``# Traverse through second array``    ``# and check if every element is``    ``# present in hash map``    ``for` `i ``in` `range` `(``len``(arr2)):``        ``x ``=` `arr2[i]` `        ``# If element is not present``        ``# in hash map or element``        ``# is not present less number``        ``# of times``        ``if` `x ``not` `in` `hM ``or` `hM[x] ``=``=` `0``:``             ``return` `False` `        ``hM[x] ``-``=` `1``       ` `    ``return` `True` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``  ` `    ``arr1 ``=` `[``2``, ``1``, ``3``, ``5``, ``4``, ``3``, ``2``]``    ``arr2 ``=` `[``3``, ``2``, ``2``, ``4``, ``5``, ``3``, ``1``]``    ``if` `(arePermutations(arr1, arr2)):``        ``print``(``"Arrays are permutations of each other"``)``    ``else``:``        ``print``(``"Arrays are NOT permutations of each other"``)``        ` `# This code is contributed by Chitranayal`

## C#

 `// C# program to find one array``// is permutation of other array``using` `System;``using` `System.Collections.Generic;` `public` `class` `Permutations {``    ``// Returns true if arr1[] and arr2[]``    ``// are permutations of each other``    ``static` `Boolean arePermutations(``int``[] arr1, ``int``[] arr2)``    ``{``        ``// Arrays cannot be permutations of one another if``        ``// they are not the same length``        ``if` `(arr1.Length != arr2.Length)``            ``return` `false``;` `        ``// Creates an empty hashMap hM``        ``Dictionary<``int``, ``int``> hM = ``new` `Dictionary<``int``, ``int``>();` `        ``// Traverse through the first array``        ``// and add elements to hash map``        ``for` `(``int` `i = 0; i < arr1.Length; i++) {``            ``int` `x = arr1[i];``            ``if` `(!hM.ContainsKey(x))``                ``hM.Add(x, 1);``            ``else` `{``                ``int` `k = hM[x];``                ``hM.Remove(x);``                ``hM.Add(x, k + 1);``            ``}``        ``}` `        ``// Traverse through second array and check if every``        ``// element is present in hash map (and the same``        ``// number of times)``        ``for` `(``int` `i = 0; i < arr2.Length; i++) {``            ``int` `x = arr2[i];` `            ``// If element is not present in hash map or``            ``// element is not present the same number of``            ``// times``            ``if` `(!hM.ContainsKey(x))``                ``return` `false``; ``// Not present in the hash map` `            ``int` `k = hM[x];``            ``if` `(k == 0)``                ``return` `false``; ``// Not present the same number``                              ``// of times``            ``hM.Remove(x);``            ``hM.Add(x, k - 1);``        ``}``        ``return` `true``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr1 = { 2, 1, 3, 5, 4, 3, 2 };``        ``int``[] arr2 = { 3, 2, 2, 4, 5, 3, 1 };``        ``if` `(arePermutations(arr1, arr2))``            ``Console.WriteLine(``"Arrays are permutations of each other"``);``        ``else``            ``Console.WriteLine(``"Arrays are NOT permutations of each other"``);``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

`Arrays are permutations of each other`

Time complexity: O(n) under the assumption that we have a hash function that inserts and finds elements in O(1) time.
Auxiliary space: O(n) because it is using map

Approach 2: No Extra Space:

Another approach to check if one array is a permutation of another array is to sort both arrays and then compare each element of both arrays. If all the elements are the same in both arrays, then they are permutations of each other. Note that the space complexity will be optimized since it does not require any extra data structure to store values.

Here’s the code for this approach:

## C++

 `#include ``using` `namespace` `std;` `// Returns true if arr1[] and arr2[] are permutations of each other``bool` `arePermutations(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m)``{``    ``// Arrays cannot be permutations of one another unless they are``    ``// of the same length``    ``if` `(n != m)``    ``{``        ``return` `false``;``    ``}` `    ``// Sort both arrays``    ``sort(arr1, arr1 + n);``    ``sort(arr2, arr2 + m);` `    ``// Compare each element of both arrays``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(arr1[i] != arr2[i])``        ``{``            ``return` `false``;``        ``}``    ``}` `    ``return` `true``;``}` `// Driver function to test above function``int` `main()``{``    ``int` `arr1[] = {2, 1, 3, 5, 4, 3, 2};``    ``int` `arr2[] = {3, 2, 2, 4, 5, 3, 1};``    ``int` `n = ``sizeof``(arr1) / ``sizeof``(arr1[0]);``    ``int` `m = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` `    ``if` `(arePermutations(arr1, arr2, n, m))``    ``{``        ``cout << ``"Arrays are permutations of each other"` `<< endl;``    ``}``    ``else``    ``{``        ``cout << ``"Arrays are NOT permutations of each other"` `<< endl;``    ``}` `    ``return` `0;``}`

## Java

 `import` `java.util.Arrays;` `public` `class` `PermutationChecker {``    ``public` `static` `boolean` `arePermutations(``int``[] arr1, ``int``[] arr2, ``int` `n, ``int` `m) {``        ``// Arrays cannot be permutations of one another unless they are``        ``// of the same length``        ``if` `(n != m) {``            ``return` `false``;``        ``}` `        ``// Sort both arrays``        ``Arrays.sort(arr1);``        ``Arrays.sort(arr2);` `        ``// Compare each element of both arrays``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(arr1[i] != arr2[i]) {``                ``return` `false``;``            ``}``        ``}` `        ``return` `true``;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int``[] arr1 = {``2``, ``1``, ``3``, ``5``, ``4``, ``3``, ``2``};``        ``int``[] arr2 = {``3``, ``2``, ``2``, ``4``, ``5``, ``3``, ``1``};``        ``int` `n = arr1.length;``        ``int` `m = arr2.length;` `        ``if` `(arePermutations(arr1, arr2, n, m)) {``            ``System.out.println(``"Arrays are permutations of each other"``);``        ``} ``else` `{``            ``System.out.println(``"Arrays are NOT permutations of each other"``);``        ``}``    ``}``}`

## Python3

 `def` `are_permutations(arr1, arr2):``    ``# Arrays cannot be permutations of one another unless they are``    ``# of the same length``    ``n ``=` `len``(arr1)``    ``m ``=` `len``(arr2)``    ``if` `n !``=` `m:``        ``return` `False``    ` `    ``# Sort both arrays``    ``arr1.sort()``    ``arr2.sort()``    ` `    ``# Compare each element of both arrays``    ``for` `i ``in` `range``(n):``        ``if` `arr1[i] !``=` `arr2[i]:``            ``return` `False``    ` `    ``return` `True` `arr1 ``=` `[``2``, ``1``, ``3``, ``5``, ``4``, ``3``, ``2``]``arr2 ``=` `[``3``, ``2``, ``2``, ``4``, ``5``, ``3``, ``1``]` `if` `are_permutations(arr1, arr2):``    ``print``(``"Arrays are permutations of each other"``)``else``:``    ``print``(``"Arrays are NOT permutations of each other"``)`

## C#

 `using` `System;` `class` `PermutationChecker {``    ``static` `bool` `ArePermutations(``int``[] arr1, ``int``[] arr2) {``        ``// Arrays cannot be permutations of one another unless they are``        ``// of the same length``        ``int` `n = arr1.Length;``        ``int` `m = arr2.Length;``        ``if` `(n != m) {``            ``return` `false``;``        ``}``        ` `        ``// Sort both arrays``        ``Array.Sort(arr1);``        ``Array.Sort(arr2);``        ` `        ``// Compare each element of both arrays``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(arr1[i] != arr2[i]) {``                ``return` `false``;``            ``}``        ``}``        ` `        ``return` `true``;``    ``}` `    ``static` `void` `Main() {``        ``int``[] arr1 = {2, 1, 3, 5, 4, 3, 2};``        ``int``[] arr2 = {3, 2, 2, 4, 5, 3, 1};` `        ``if` `(ArePermutations(arr1, arr2)) {``            ``Console.WriteLine(``"Arrays are permutations of each other"``);``        ``} ``else` `{``            ``Console.WriteLine(``"Arrays are NOT permutations of each other"``);``        ``}``    ``}``}`

## Javascript

 `function` `arePermutations(arr1, arr2) {``  ``// Arrays cannot be permutations of one another unless they are``  ``// of the same length``  ``const n = arr1.length;``  ``const m = arr2.length;``  ``if` `(n !== m) {``    ``return` `false``;``  ``}` `  ``// Sort both arrays``  ``arr1.sort();``  ``arr2.sort();` `  ``// Compare each element of both arrays``  ``for` `(let i = 0; i < n; i++) {``    ``if` `(arr1[i] !== arr2[i]) {``      ``return` `false``;``    ``}``  ``}` `  ``return` `true``;``}` `const arr1 = [2, 1, 3, 5, 4, 3, 2];``const arr2 = [3, 2, 2, 4, 5, 3, 1];` `if` `(arePermutations(arr1, arr2)) {``  ``console.log(``"Arrays are permutations of each other"``);``} ``else` `{``  ``console.log(``"Arrays are NOT permutations of each other"``);``}`

Output

`Arrays are permutations of each other`

Time complexity: O(N*log(N)), Where N is the size of the arrays
Auxiliary space: O(1)

Approach: Using DFS

1. We start by defining a helper function called dfs that performs the DFS traversal. This function takes the following parameters:
• arr1 and arr2: The two arrays we want to check for permutations.
• visited: A boolean array to keep track of the elements in arr2 that have been used.
• index: The current index in arr1 that we are matching.
1. The base case of the DFS function is when we have checked all elements in arr1. If the index reaches the length of arr1, it means we have successfully matched all elements, so we return true.
2. For the current element in arr1 at index index, we iterate over arr2 to find a matching element that hasn’t been used before. If we find a match, we mark the element in arr2 as visited and recursively call dfs with the next index.
3. If the recursive call returns true, it means a permutation has been found, so we return true from the current call as well. Otherwise, we backtrack by marking the element in arr2 as not visited and continue searching for other permutations.
4. The arePermutations function is the entry point of the algorithm. It takes the two arrays arr1 and arr2 as input.
5. First, we check if the lengths of the arrays arr1 and arr2 are different. If they are not equal, the arrays cannot be permutations of each other, so we return false.
6. Next, we initialize a boolean visited array with false values to keep track of the used elements in arr2.
7. We then call the dfs function, passing in arr1, arr2, visited, and the starting index of 0.
8. If the dfs function returns true, it means a permutation has been found, so we return true from the arePermutations function as well.
9. If no permutation is found after the DFS traversal, we return false.

By using DFS, the modified code explores all possible paths in the search space, which corresponds to the different permutations of the arrays. The backtracking step allows the algorithm to efficiently search for permutations by avoiding unnecessary exploration of paths that cannot lead to valid permutations.

## C++

 `#include ``using` `namespace` `std;` `// Helper function to perform DFS``bool` `dfs(vector<``int``>& arr1, vector<``int``>& arr2, vector<``bool``>& visited, ``int` `index) {``    ``// Base case: All elements have been visited``    ``if` `(index == arr1.size()) {``        ``return` `true``;``    ``}` `    ``// Check if the current element in arr1 has a corresponding element in arr2``    ``int` `num = arr1[index];``    ``for` `(``int` `i = 0; i < arr2.size(); i++) {``        ``if` `(!visited[i] && arr2[i] == num) {``            ``visited[i] = ``true``;``            ``if` `(dfs(arr1, arr2, visited, index + 1)) {``                ``return` `true``;``            ``}``            ``visited[i] = ``false``; ``// Backtrack``        ``}``    ``}` `    ``return` `false``;``}` `// Returns true if arr1[] and arr2[] are permutations of each other``bool` `arePermutations(vector<``int``>& arr1, vector<``int``>& arr2) {``    ``int` `n = arr1.size();``    ``int` `m = arr2.size();` `    ``// Arrays cannot be permutations of one another unless they are of the same length``    ``if` `(n != m) {``        ``return` `false``;``    ``}` `    ``// Initialize a visited array to keep track of used elements in arr2``    ``vector<``bool``> visited(m, ``false``);` `    ``// Perform DFS to check for permutations``    ``return` `dfs(arr1, arr2, visited, 0);``}` `// Driver function to test above function``int` `main() {``    ``vector<``int``> arr1 = {2, 1, 3, 5, 4, 3, 2};``    ``vector<``int``> arr2 = {3, 2, 2, 4, 5, 3, 1};` `    ``if` `(arePermutations(arr1, arr2)) {``        ``cout << ``"Arrays are permutations of each other"` `<< endl;``    ``} ``else` `{``        ``cout << ``"Arrays are NOT permutations of each other"` `<< endl;``    ``}` `    ``return` `0;``}`

Output

```Arrays are permutations of each other
```

Time complexity: O(n log n) ,where n is the length of the arrays.

Auxiliary space: O(n), where n is the length of the arrays.

My Personal Notes arrow_drop_up