Check if two arrays are equal or not
Given two given arrays of equal length, the task is to find if 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 permutation) of elements may be different though.
Note: 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 : Yes
Input : arr1[] = {1, 2, 5, 4, 0, 2, 1};
arr2[] = {2, 4, 5, 0, 1, 1, 2};
Output : Yes
Input : arr1[] = {1, 7, 1};
arr2[] = {7, 7, 1};
Output : No
A simple solution is to sort both arrays and then linearly compare elements.
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 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); if (areEqual(arr1, arr2, n, m)) cout << "Yes"; else cout << "No"; return 0;} |
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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 }; if (areEqual(arr1, arr2)) System.out.println("Yes"); else System.out.println("No"); }} |
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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 - 1): if (arr1[i] != arr2[i]): return False # If all elements were same. return True# Driver Codearr1 = [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")# This code is contributed# by Shivi_Aggarwal. |
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C#
// C# program to find given two array// are equal or notusing 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 code public static void Main() { int[] arr1 = { 3, 5, 2, 5, 2 }; int[] arr2 = { 2, 3, 5, 5, 2 }; if (areEqual(arr1, arr2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by anuj_67. |
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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);if (areEqual($arr1, $arr2, $n, $m)) echo "Yes";else echo "No";// This code is contributed by anuj_67.?> |
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Output
Yes
Time Complexity: O(n log n)
Auxiliary Space: O(1)
An Efficient Solution to this approach is to use hashing. We store all elements of arr1[] and their counts in a hash table. Then we traverse arr2[] and check if the count of every element in arr2[] matches with the count in arr1[].
Below is the implementation of the above idea. We use unordered_map to store counts.
C++
// C++ program to find given two array// are equal or not using hashing technique#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; mp[arr2[i]]--; } return true;}// Driver 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); if (areEqual(arr1, arr2, n, m)) cout << "Yes"; else cout << "No"; return 0;} |
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Java
// Java program to find given two array// are equal or not using hashing techniqueimport java.util.*;import java.io.*;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 code public static void main(String[] args) { int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; if (areEqual(arr1, arr2)) System.out.println("Yes"); else System.out.println("No"); }} |
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Python3
# Python3 program to find if given# two arrays are equal or not# using dictionaryfrom collections import defaultdict# 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 # Create a defaultdict count to # store counts count = defaultdict(int) # Store the elements of arr1 # and their counts in the dictionary for i in arr1: 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 elemnent # is not in arr2 or if any element # appears more no. of times than in arr1 if (count[i] == 0): return False # If element is found, decrement # its value in the dictionary else: count[i] -= 1 # Return true if both arr1 and # arr2 are equal return True# Driver Codearr1 = [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")# This code is contributed by Karthik_Aravind |
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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 code public static void Main(String[] args) { int[] arr1 = { 3, 5, 2, 5, 2 }; int[] arr2 = { 2, 3, 5, 5, 2 }; if (areEqual(arr1, arr2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}/* This code contributed by PrinciRaj1992 */ |
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Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(n)
An Alternate Solution without comparing each element of the arrays and without using unordered_map (by using XOR)
C++14
// 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; // to store xor of both arrays int b1 = arr1[0]; int b2 = arr2[0]; // find xor of each elements in array for (int i = 1; i < n; i++) { b1 ^= arr1[i]; } for (int i = 1; i < m; i++) { b2 ^= arr2[i]; } int all_xor = b1 ^ b2; // if xor is zero means they are equal (5^5=0) if (all_xor == 0) return true; // If all elements were not same, then xor will not be // zero return false;}// Driver 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;} |
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Java
// Java program to find given two array// are equal or not import 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[]){ // Length of the two array int n = arr1.length; int m = arr2.length; // If lengths of arrays are not equal if (n != m) return false; // To store xor of both arrays int b1 = arr1[0]; int b2 = arr2[0]; // Find xor of each elements in array for(int i = 1; i < n; i++) { b1 ^= arr1[i]; } for(int i = 1; i < m; i++) { b2 ^= arr2[i]; } int all_xor = b1 ^ b2; // If xor is zero means they are // equal (5^5=0) if (all_xor == 0) return true; // If all elements were not same, // then xor will not be zero return false;}// Driver codepublic 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");}}// This code is contributed by sayantanbose2001 |
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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 b1 = arr1[0] b2 = arr2[0] # find xor of all elements for i in range(1, n - 1): b1 ^= arr1[i] for i in range(1, m - 1): b2 ^= arr2[i] all_xor = b1 ^ b2 # If all elements were same then xor will be zero if(all_xor == 0): return True return False# Driver Codearr1 = [3, 5, 2, 5, 2]arr2 = [2, 3, 5, 5, 2]n = len(arr1)m = len(arr2)# Function callif (areEqual(arr1, arr2, n, m)): print("Yes")else: print("No") |
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Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(1)
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Improved By: vt_m, Shivi_Aggarwal, imrohan, princiraj1992, karthikaravindt88, anupriyanishad
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