Check if two arrays are equal or not
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 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 ); // 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 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 }; // 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 Code if __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 approach 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'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. |
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 call if (areEqual( $arr1 , $arr2 , $N , $M )) echo "Yes" ; else echo "No" ; ?> |
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> |
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 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 ); // Function call if (areEqual(arr1, arr2, N, M)) cout << "Yes" ; else cout << "No" ; return 0; } |
Java
// Java program for the above approach 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 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 Code if __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 technique using 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> |
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)
Check if two arrays are equal or not using Map
- Initialise a map say unmap
- Insert all elements of array A into map
- Remove all elements of array B from map
- Check if the size of unmap becomes zero
- If zero, return true
- Otherwise, return false
Below is the implementation of the above approach:
C++
// CPP program to find Non-overlapping sum #include <bits/stdc++.h> using namespace std; // function for calculating // Non-overlapping sum of two array bool isEqual(vector< int > A, vector< int > B) { unordered_map< int , int > unmap; // Insert all element of A into map for ( auto i : A) unmap[i]++; // Remove all elements of B from map for ( auto i : B) { unmap[i]--; if (unmap[i] == 0) unmap.erase(i); } // Check if size of unmap becomes zero // If zero, return true // Otherwise, return false if (unmap.size() == 0) return true ; return false ; } // Driver code int main() { vector< int > A = { 5, 4, 9, 2, 3 }; vector< int > B = { 5, 4, 9, 2, 3 }; // function call cout << isEqual(A, B); return 0; } // This code is contributed by hkdass001 |
Java
// Java program to find Non-overlapping sum import java.util.*; public class GFG { // function for calculating // Non-overlapping sum of two array static boolean isEqual( int [] A, int [] B) { HashMap<Integer, Integer> unmap = new HashMap<>(); // Insert all element of A into map for ( int i : A) unmap.put(i, unmap.getOrDefault(i, 0 ) + 1 ); // Remove all elements of B from map for ( int i : B) { if (unmap.containsKey(i)) { unmap.put(i, unmap.get(i) - 1 ); if (unmap.get(i) == 0 ) unmap.remove(i); } } // Check if size of unmap becomes zero // If zero, return true // Otherwise, return false if (unmap.size() == 0 ) return true ; return false ; } // Driver code public static void main(String[] args) { int [] A = { 5 , 4 , 9 , 2 , 3 }; int [] B = { 5 , 4 , 9 , 2 , 3 }; // function call System.out.println((isEqual(A, B) ? "1" : "0" )); } } // This code is contributed by Karandeep1234 |
Python3
# Python program to find Non-overlapping sum from collections import Counter # function for calculating # Non-overlapping sum of two array def isEqual(A, B): # Use Counter to count the frequency of elements unmap = Counter(A) # Remove all elements of B from map for i in B: unmap[i] - = 1 if unmap[i] = = 0 : del unmap[i] # Check if size of unmap becomes zero # If zero, return true # Otherwise, return false if len (unmap) = = 0 : return True return False # Driver code A = [ 5 , 4 , 9 , 2 , 3 ] B = [ 5 , 4 , 9 , 2 , 3 ] # function call #rint(isEqual(A, B)) print ( 1 if isEqual(A, B) else 0 ) |
C#
// C# program to find Non-overlapping sum using System; using System.Collections.Generic; class Program { static bool IsEqual(List< int > A, List< int > B) { Dictionary< int , int > unmap = new Dictionary< int , int >(); foreach ( int i in A) { if (unmap.ContainsKey(i)) { unmap[i]++; } else { unmap[i] = 1; } } foreach ( int i in B) { if (unmap.ContainsKey(i)) { unmap[i]--; if (unmap[i] == 0) { unmap.Remove(i); } } } if (unmap.Count == 0) { return true ; } return false ; } static void Main( string [] args) { List< int > A = new List< int >{ 5, 4, 9, 2, 3 }; List< int > B = new List< int >{ 5, 4, 9, 2, 3 }; Console.WriteLine(IsEqual(A, B) ? 1 : 0); } } // This code is contributed by Susobhan Akhuli |
Javascript
// Javascript Function function isEqual(A, B) { let unmap = {}; // Insert all elements of A into map for (let i of A) { unmap[i] = (unmap[i] || 0) + 1; } // Remove all elements of B from map for (let i of B) { if (unmap[i]) { unmap[i]--; if (unmap[i] === 0) { delete unmap[i]; } } } // Check if size of unmap becomes zero // If zero, return true // Otherwise, return false if (Object.keys(unmap).length === 0) { return true ; } return false ; } // Driver Code let A = [5, 4, 9, 2, 3]; let B = [5, 4, 9, 2, 3]; // function call console.log(isEqual(A, B)); // This code is contributed by akashish__ |
1
Time Complexity: O(N)
Auxiliary Space: O(N)
Check if two arrays are equal or not using Counter Class
We can use the Counter class from the collections module to count the number of occurrences of each element in the arrays and then compare the resulting dictionaries.
Steps:
- Use the Counter class to count the number of occurrences of each element in a and b.
- Use the == operator to compare the resulting Counter objects.
- Return the result of the comparison.
- Import the Counter class from the collections module.
Below is the implementation of the above approach:
C++
#include <map> #include <iostream> #include <algorithm> bool arrays_equal(std::vector< int > a, std::vector< int > b) { std::map< int , int > map_a, map_b; for ( auto i : a) { map_a[i]++; } for ( auto i : b) { map_b[i]++; } return map_a == map_b; } int main() { std::vector< int > a = {3, 2, 1, 3, 2, 1}; std::vector< int > b = {1, 2, 3, 1, 2, 3}; std::vector< int > c = {4, 5, 6, 4, 5, 6}; std::cout << (arrays_equal(a, b) ? "True" : "False" ) << std::endl; std::cout << (arrays_equal(a, c) ? "True" : "False" ) << std::endl; return 0; } |
Java
import java.util.*; public class GFG { public static boolean arrays_equal( int [] a, int [] b) { Map<Integer, Integer> aCount = new HashMap<>(); Map<Integer, Integer> bCount = new HashMap<>(); for ( int i = 0 ; i < a.length; i++) { int count = aCount.getOrDefault(a[i], 0 ); aCount.put(a[i], count + 1 ); } for ( int i = 0 ; i < b.length; i++) { int count = bCount.getOrDefault(b[i], 0 ); bCount.put(b[i], count + 1 ); } return aCount.equals(bCount); } public static void main(String[] args) { int [] a = { 3 , 2 , 1 , 3 , 2 , 1 }; int [] b = { 1 , 2 , 3 , 1 , 2 , 3 }; int [] c = { 4 , 5 , 6 , 4 , 5 , 6 }; System.out.println( arrays_equal(a, b) ? "True" : "False" ); // True System.out.println( arrays_equal(a, c) ? "True" : "False" ); // False } } // This code is contributed by Susobhan Akhuli |
Python3
from collections import Counter def arrays_equal(a, b): return Counter(a) = = Counter(b) a = [ 3 , 2 , 1 , 3 , 2 , 1 ] b = [ 1 , 2 , 3 , 1 , 2 , 3 ] c = [ 4 , 5 , 6 , 4 , 5 , 6 ] print (arrays_equal(a, b)) # True print (arrays_equal(a, c)) # False # This code is contributed by Susobhan Akhuli |
C#
// C# implemenatation of above approach. using System; using System.Collections.Generic; using System.Linq; public class gfg { // Method to check if two arrays have the same set of elements, ignoring order static bool ArraysEqual(List< int > a, List< int > b) { // Sort both arrays and compare them using the SequenceEqual method return a.OrderBy(x => x).SequenceEqual(b.OrderBy(x => x)); } // Main method to test the ArraysEqual method public static void Main() { // Initialize three lists of integers List< int > a = new List< int > { 3, 2, 1, 3, 2, 1 }; List< int > b = new List< int > { 1, 2, 3, 1, 2, 3 }; List< int > c = new List< int > { 4, 5, 6, 4, 5, 6 }; // Test the ArraysEqual method with the three lists Console.WriteLine(ArraysEqual(a, b) ? "True" : "False" ); Console.WriteLine(ArraysEqual(a, c) ? "True" : "False" ); } } // This code is contributed by Amit Mangal. |
Javascript
// JavaScript implemenatation of above approach. // Function to check if two arrays have the same set of elements, ignoring order function arraysEqual(a, b) { // Sort the arrays and convert them to strings, then compare the strings return JSON.stringify(a.slice().sort()) === JSON.stringify(b.slice().sort()); } // Declare three arrays const a = [3, 2, 1, 3, 2, 1]; const b = [1, 2, 3, 1, 2, 3]; const c = [4, 5, 6, 4, 5, 6]; // Output the result of arraysEqual with the two pairs of arrays console.log(arraysEqual(a, b) ? "True" : "False" ); console.log(arraysEqual(a, c) ? "True" : "False" ); // This code is contributed by Amit Mangal. |
True False
Time Complexity: O(N) [To count the number of occurrences of each element in the arrays]
Auxiliary Space: O(1)
Check if two arrays are equal or not using using Priority Queue
We can create a custom function to compare two arrays using priority queues.
Steps:
- First inserting all elements of both the arrays into their respective priority queues
- Then looping through them and comparing each element in the arrays.
- If all elements are equal, then return true means, the two arrays are equal.
- Otherwise 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; // Function to check if two arrays are equal or not. bool arrays_equal(vector< int > A, vector< int > B, int N, int M) { // If lengths of array are not equal means // array are not equal if (N != M) return false ; // Creating 2 priority queues for 2 arrays priority_queue< int > pq1; priority_queue< int > pq2; // Insert elements of both the arrays into their // respective priority queues for ( int i = 0; i < N; i++) { pq1.push(A[i]); pq2.push(B[i]); } // Traverse and check if the top elements of both the // priority queues are same or not while (N--) { // If top elements are not same then return false if (pq1.top() != pq2.top()) return false ; // If top elements are same then // Popping top elements of both the priority queues pq1.pop(); pq2.pop(); } // If all elements were same. return true ; } // Driver's Code int main() { vector< int > arr1 = { 3, 2, 1, 3, 2, 1 }; vector< int > arr2 = { 1, 2, 3, 1, 2, 3 }; vector< int > arr3 = { 4, 5, 6, 4, 5, 6 }; int M = sizeof (arr1) / sizeof ( int ); int N = sizeof (arr2) / sizeof ( int ); int L = sizeof (arr3) / sizeof ( int ); cout << (arrays_equal(arr1, arr2, M, N) ? "True" : "False" ) << endl; cout << (arrays_equal(arr1, arr3, M, L) ? "True" : "False" ) << endl; return 0; } // This code is contributed by Susobhan Akhuli |
Python3
import heapq def arrays_equal(A, B): # If lengths of array are not equal means array are not equal if len (A) ! = len (B): return False # Creating 2 heaps for 2 arrays heap1 = list (A) heap2 = list (B) # Convert list into heap heapq.heapify(heap1) heapq.heapify(heap2) # Traverse and check if the top elements of both the # heaps are same or not while heap1: # If top elements are not same then return false if heapq.heappop(heap1) ! = heapq.heappop(heap2): return False # If all elements were same. return True # Example usage arr1 = [ 3 , 2 , 1 , 3 , 2 , 1 ] arr2 = [ 1 , 2 , 3 , 1 , 2 , 3 ] arr3 = [ 4 , 5 , 6 , 4 , 5 , 6 ] print (arrays_equal(arr1, arr2)) # Output: True print (arrays_equal(arr1, arr3)) # Output: False |
Java
import java.util.*; public class ArraysEqual { // Function to check if two arrays are equal or not. public static boolean arraysEqual( int [] A, int [] B, int N, int M) { // If lengths of array are not equal means // array are not equal if (N != M) return false ; // Creating 2 priority queues for 2 arrays PriorityQueue<Integer> pq1 = new PriorityQueue<Integer>( Collections.reverseOrder()); PriorityQueue<Integer> pq2 = new PriorityQueue<Integer>( Collections.reverseOrder()); // Insert elements of both the arrays into their // respective priority queues for ( int i = 0 ; i < N; i++) { pq1.offer(A[i]); pq2.offer(B[i]); } // Traverse and check if the top elements of both // the priority queues are same or not while (N-- > 0 ) { // If top elements are not same then return // false if (!pq1.peek().equals(pq2.peek())) return false ; // If top elements are same then // Popping top elements of both the priority // queues pq1.poll(); pq2.poll(); } // If all elements were same. return true ; } // Driver's Code public static void main(String[] args) { int [] arr1 = { 3 , 2 , 1 , 3 , 2 , 1 }; int [] arr2 = { 1 , 2 , 3 , 1 , 2 , 3 }; int [] arr3 = { 4 , 5 , 6 , 4 , 5 , 6 }; int M = arr1.length; int N = arr2.length; int L = arr3.length; System.out.println(arraysEqual(arr1, arr2, M, N) ? "True" : "False" ); System.out.println(arraysEqual(arr1, arr3, M, L) ? "True" : "False" ); } } |
Javascript
function arraysEqual(A, B) { // If lengths of array are not equal means array are not equal if (A.length !== B.length) { return false ; } // Creating 2 heaps for 2 arrays const heap1 = [...A]; const heap2 = [...B]; // Convert list into heap heap1.sort((a, b) => a - b); heap2.sort((a, b) => a - b); // Traverse and check if the top elements of both the // heaps are same or not while (heap1.length > 0) { // If top elements are not same then return false if (heap1.shift() !== heap2.shift()) { return false ; } } // If all elements were same. return true ; } // Example usage const arr1 = [3, 2, 1, 3, 2, 1]; const arr2 = [1, 2, 3, 1, 2, 3]; const arr3 = [4, 5, 6, 4, 5, 6]; console.log(arraysEqual(arr1, arr2)); // Output: true console.log(arraysEqual(arr1, arr3)); // Output: false |
C#
// GFG // C# code for the given approach using System; using System.Linq; class Program { static bool ArraysEqual( int [] A, int [] B) { // If lengths of array are not equal means // array are not equal if (A.Length != B.Length) return false ; // Sort the two input arrays Array.Sort(A); Array.Sort(B); // Traverse both arrays and check if their elements // are equal for ( int i = 0; i < A.Length; i++) { if (A[i] != B[i]) return false ; } // If all elements are equal return true ; } static void Main( string [] args) { int [] arr1 = { 3, 2, 1, 3, 2, 1 }; int [] arr2 = { 1, 2, 3, 1, 2, 3 }; int [] arr3 = { 4, 5, 6, 4, 5, 6 }; Console.WriteLine(ArraysEqual(arr1, arr2) ? "True" : "False" ); // True Console.WriteLine(ArraysEqual(arr1, arr3) ? "True" : "False" ); // False } } // This code is written by Sundaram |
True False
Time Complexity: O(N), where N is the size of the arrays.
Auxiliary Space: O(N) [To store elements into priority queues]
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