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.

- Create a Hash Map for all the elements of arr1[] such that array elements are keys and their counts are values.
- 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.
- If all elements are found then return true.

Below is the implementation of this approach.

## Java

`// A Java program to find one array is permutation of other array` `import` `java.util.HashMap;` `class` `Permutaions` `{` ` ` `// 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<Integer, Integer> hM = ` `new` `HashMap<Integer, Integer>();` ` ` `// 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"` `);` ` ` `}` `}` |

*chevron_right*

*filter_none*

## 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` |

*chevron_right*

*filter_none*

## C#

`// C# program to find one array` `// is permutation of other array` `using` `System;` `using` `System.Collections.Generic;` `public` `class` `Permutaions {` ` ` `// 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 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 */` |

*chevron_right*

*filter_none*

**Output**

Arrays are permutations of each other

The time complexity of this method is O(n) under the assumption that we have a hash function inserts and finds elements in O(1) time.

This article is contributed by **Ravi**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important **Java Foundation** and Collections concepts with the **Fundamentals of Java and Java Collections Course** at a student-friendly price and become industry ready.