Given two arrays, arr1 and arr2 of equal length N, the task is to find if the given arrays are equal or not.
Two arrays are said to be equal if:
- both of them contain the same set of elements,
- arrangements (or permutations) of elements might/might not be same.
- If there are repetitions, then counts of repeated elements must also be the same for two arrays to be equal.
Examples:
Input: arr1[] = {1, 2, 5, 4, 0}, arr2[] = {2, 4, 5, 0, 1}
Output: YesInput: arr1[] = {1, 2, 5, 4, 0, 2, 1}, arr2[] = {2, 4, 5, 0, 1, 1, 2}
Output: YesInput: arr1[] = {1, 7, 1}, arr2[] = {7, 7, 1}
Output: No
Check if two arrays are equal or not using Sorting
Follow the steps below to solve the problem using this approach:
- Sort both the arrays
- Then linearly compare elements of both the arrays
- If all are equal then return true, else return false
Below is the implementation of the above approach:
C++
// C++ program to find given two array// are equal or not#include <bits/stdc++.h>using namespace std;// Returns true if arr1[0..n-1] and arr2[0..m-1]// contain same elements.bool areEqual(int arr1[], int arr2[], int N, int M){ // If lengths of array are not equal means // array are not equal if (N != M) return false; // Sort both arrays sort(arr1, arr1 + N); sort(arr2, arr2 + M); // Linearly compare elements for (int i = 0; i < N; i++) if (arr1[i] != arr2[i]) return false; // If all elements were same. return true;}// Driver's Codeint main(){ int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; int N = sizeof(arr1) / sizeof(int); int M = sizeof(arr2) / sizeof(int); // Function call if (areEqual(arr1, arr2, N, M)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to find given two array// are equal or notimport java.io.*;import java.util.*;class GFG { // Returns true if arr1[0..n-1] and // arr2[0..m-1] contain same elements. public static boolean areEqual(int arr1[], int arr2[]) { int N = arr1.length; int M = arr2.length; // If lengths of array are not equal means // array are not equal if (N != M) return false; // Sort both arrays Arrays.sort(arr1); Arrays.sort(arr2); // Linearly compare elements for (int i = 0; i < N; i++) if (arr1[i] != arr2[i]) return false; // If all elements were same. return true; } // Driver code public static void main(String[] args) { int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; // Function call if (areEqual(arr1, arr2)) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# Python3 program to find given# two array are equal or not# Returns true if arr1[0..n-1] and# arr2[0..m-1] contain same elements.def areEqual(arr1, arr2, N, M): # If lengths of array are not # equal means array are not equal if (N != M): return False # Sort both arrays arr1.sort() arr2.sort() # Linearly compare elements for i in range(0, N): if (arr1[i] != arr2[i]): return False # If all elements were same. return True# Driver Codeif __name__ == "__main__": arr1 = [3, 5, 2, 5, 2] arr2 = [2, 3, 5, 5, 2] n = len(arr1) m = len(arr2) if (areEqual(arr1, arr2, n, m)): print("Yes") else: print("No") |
C#
// C# program for the above approachusing System;class GFG { // Returns true if arr1[0..n-1] and // arr2[0..m-1] contain same elements. public static bool areEqual(int[] arr1, int[] arr2) { int N = arr1.Length; int M = arr2.Length; // If lengths of array are not // equal means array are not equal if (N != M) return false; // Sort both arrays Array.Sort(arr1); Array.Sort(arr2); // Linearly compare elements for (int i = 0; i < N; i++) if (arr1[i] != arr2[i]) return false; // If all elements were same. return true; } // Driver's code public static void Main() { int[] arr1 = { 3, 5, 2, 5, 2 }; int[] arr2 = { 2, 3, 5, 5, 2 }; // Function call if (areEqual(arr1, arr2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by anuj_67. |
Javascript
<script>// JavaScript program for the above approach // Returns true if arr1[0..n-1] and arr2[0..m-1] // contain same elements. function areEqual(arr1, arr2) { let N = arr1.length; let M = arr2.length; // If lengths of array are not equal means // array are not equal if (N != M) return false; // Sort both arrays arr1.sort(); arr2.sort(); // Linearly compare elements for (let i = 0; i < N; i++) if (arr1[i] != arr2[i]) return false; // If all elements were same. return true; }// Driver Code let arr1 = [3, 5, 2, 5, 2]; let arr2 = [2, 3, 5, 5, 2]; if (areEqual(arr1, arr2)) document.write("Yes"); else document.write("No");</script> |
PHP
<?php// PHP program to find given // two array are equal or not// Returns true if arr1[0..n-1]// and arr2[0..m-1] contain same elements.function areEqual( $arr1, $arr2, $N, $M){ // If lengths of array // are not equal means // array are not equal if ($N != $M) return false; // Sort both arrays sort($arr1); sort($arr2); // Linearly compare elements for ( $i = 0; $i < $N; $i++) if ($arr1[$i] != $arr2[$i]) return false; // If all elements were same. return true;}// Driver Code$arr1 = array(3, 5, 2, 5, 2);$arr2 = array(2, 3, 5, 5, 2);$N = count($arr1);$M = count($arr2);// Function callif (areEqual($arr1, $arr2, $N, $M)) echo "Yes";else echo "No";?> |
Output
Yes
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
Check if two arrays are equal or not using Hashing:
Store count of all elements of arr1[] in a hash table. Then traverse arr2[] and check if the count of every element in arr2[] matches with the count of elements of arr1[].
Follow the steps mentioned below to implement the approach:
- First check if length of arr1 is not equal to the length of arr2 then return false
- Then traverse over first array and store the count of every element in the hash map
- Then traverse over second array and decrease the count of its elements in the hash map. If that element is not present or the count of that element is
zero in the hash map, then return false, else decrease the count of that element - Return true at the end, as both the arrays are equal by now
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Returns true if arr1[0..N-1] and arr2[0..M-1]// contain same elements.bool areEqual(int arr1[], int arr2[], int N, int M){ // If lengths of arrays are not equal if (N != M) return false; // Store arr1[] elements and their counts in // hash map unordered_map<int, int> mp; for (int i = 0; i < N; i++) mp[arr1[i]]++; // Traverse arr2[] elements and check if all // elements of arr2[] are present same number // of times or not. for (int i = 0; i < N; i++) { // If there is an element in arr2[], but // not in arr1[] if (mp.find(arr2[i]) == mp.end()) return false; // If an element of arr2[] appears more // times than it appears in arr1[] if (mp[arr2[i]] == 0) return false; // decrease the count of arr2 elements in the // unordered map mp[arr2[i]]--; } return true;}// Driver's Codeint main(){ int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; int N = sizeof(arr1) / sizeof(int); int M = sizeof(arr2) / sizeof(int); // Function call if (areEqual(arr1, arr2, N, M)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Returns true if arr1[0..N-1] and arr2[0..M-1] // contain same elements. public static boolean areEqual(int arr1[], int arr2[]) { int N = arr1.length; int M = arr2.length; // If lengths of arrays are not equal if (N != M) return false; // Store arr1[] elements and their counts in // hash map Map<Integer, Integer> map = new HashMap<Integer, Integer>(); int count = 0; for (int i = 0; i < N; i++) { if (map.get(arr1[i]) == null) map.put(arr1[i], 1); else { count = map.get(arr1[i]); count++; map.put(arr1[i], count); } } // Traverse arr2[] elements and check if all // elements of arr2[] are present same number // of times or not. for (int i = 0; i < N; i++) { // If there is an element in arr2[], but // not in arr1[] if (!map.containsKey(arr2[i])) return false; // If an element of arr2[] appears more // times than it appears in arr1[] if (map.get(arr2[i]) == 0) return false; count = map.get(arr2[i]); --count; map.put(arr2[i], count); } return true; } // Driver's code public static void main(String[] args) { int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; // Function call if (areEqual(arr1, arr2)) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# Python3 program for the above approach# Returns true if arr1[0..N-1] and# arr2[0..M-1] contain same elements.def is_arr_equal(arr1, arr2): # Check if the length of arrays are # equal or not: A Easy Logic Check if len(arr1) != len(arr2): return False # Create a dict named count to # store counts of each element count = {} # Store the elements of arr1 # and their counts in the dictionary for i in arr1: if i in count: # Element already in dict, simply increment its count count[i] += 1 else: # Element found for first time, initialize it with value 1. count[i] = 1 # Traverse through arr2 and compare # the elements and its count with # the elements of arr1 for i in arr2: # Return false if the element # is not in count or if any element # appears more no. of times than in arr1 if i not in count or count[i] == 0: return False else: # If element is found, decrement # its value in the dictionary count[i] -= 1 # Return true if both arr1 and # arr2 are equal return True# Driver Codeif __name__ == "__main__": arr1 = [3, 5, 2, 5, 2] arr2 = [2, 3, 5, 5, 2] if is_arr_equal(arr1, arr2): print("Yes") else: print("No") |
C#
// C# program to find given two array// are equal or not using hashing techniqueusing System;using System.Collections.Generic;class GFG { // Returns true if arr1[0..N-1] and arr2[0..M-1] // contain same elements. public static bool areEqual(int[] arr1, int[] arr2) { int N = arr1.Length; int M = arr2.Length; // If lengths of arrays are not equal if (N != M) return false; // Store arr1[] elements and their counts in // hash map Dictionary<int, int> map = new Dictionary<int, int>(); int count = 0; for (int i = 0; i < N; i++) { if (!map.ContainsKey(arr1[i])) map.Add(arr1[i], 1); else { count = map[arr1[i]]; count++; map.Remove(arr1[i]); map.Add(arr1[i], count); } } // Traverse arr2[] elements and check if all // elements of arr2[] are present same number // of times or not. for (int i = 0; i < N; i++) { // If there is an element in arr2[], but // not in arr1[] if (!map.ContainsKey(arr2[i])) return false; // If an element of arr2[] appears more // times than it appears in arr1[] if (map[arr2[i]] == 0) return false; count = map[arr2[i]]; --count; if (!map.ContainsKey(arr2[i])) map.Add(arr2[i], count); } return true; } // Driver's code public static void Main(String[] args) { int[] arr1 = { 3, 5, 2, 5, 2 }; int[] arr2 = { 2, 3, 5, 5, 2 }; // Function call if (areEqual(arr1, arr2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// Javascript program to find given two array// are equal or not using hashing technique // Returns true if arr1[0..N-1] and arr2[0..M-1] // contain same elements. function areEqual(arr1, arr2) { let N = arr1.length; let M = arr2.length; // If lengths of arrays are not equal if (N != M) return false; // Store arr1[] elements and their counts in // hash map let map = new Map(); let count = 0; for (let i = 0; i < N; i++) { if (map.get(arr1[i]) == null) map.set(arr1[i], 1); else { count = map.get(arr1[i]); count++; map.set(arr1[i], count); } } // Traverse arr2[] elements and check if all // elements of arr2[] are present same number // of times or not. for (let i = 0; i < N; i++) { // If there is an element in arr2[], but // not in arr1[] if (!map.has(arr2[i])) return false; // If an element of arr2[] appears more // times than it appears in arr1[] if (map.get(arr2[i]) == 0) return false; count = map.get(arr2[i]); --count; map.set(arr2[i], count); } return true; } // Driver code let arr1 = [3, 5, 2, 5, 2]; let arr2 = [2, 3, 5, 5, 2]; // Function call if (areEqual(arr1, arr2)) document.write("Yes"); else document.write("No"); </script> |
Output
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)
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