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
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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 arrayimport 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"); }} |
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Python3
# Python3 program to find one # array is permutation of other arrayfrom collections import defaultdict# Returns true if arr1[] and # arr2[] are permutations of # each otherdef 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 codeif __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 |
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C#
// C# program to find one array// is permutation of other arrayusing 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 */ |
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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
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