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 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 same for two array 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 array and then linearly compare elements.
C/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 Code int 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; } |
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[]) { 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"); } } |
Python 3
# 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 Code 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"); # This code is contributed # by Shivi_Aggarwal. |
C#
// C# program to find given two array // are equal or not using 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. |
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. ?> |
Output :
Yes
Time Complexity : O(n log n)
Auxiliary Space : O(1)
An Efficient solution of 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 count of every element in arr2[] matches with count in arr1[].
Below is the implementation of above idea. We use unordered_map to store counts.
C/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 Code int 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; } |
Java
// Java program to find given two array // are equal or not using hashing technique import 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"); } } |
Output :
Yes
Time Complexity : O(n)
Auxiliary Space : O(n)
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