From 47948603f00b090fab46728dad51fe4898c296cf Mon Sep 17 00:00:00 2001
From: Rishab Dugar <56007479+kingrishabdugar@users.noreply.github.com>
Date: Sat, 8 Oct 2022 10:07:33 +0530
Subject: [PATCH] Create Efficient - Program to check if two Arrays are Equal
 or not

Using Hashing
Time Complexity: O(N)
Auxiliary Space: O(N)
---
 ...am to check if two Arrays are Equal or not | 69 +++++++++++++++++++
 1 file changed, 69 insertions(+)
 create mode 100644 Efficient - Program to check if two Arrays are Equal or not

diff --git a/Efficient - Program to check if two Arrays are Equal or not b/Efficient - Program to check if two Arrays are Equal or not
new file mode 100644
index 0000000..cf0ae3c
--- /dev/null
+++ b/Efficient - Program to check if two Arrays are Equal or not	
@@ -0,0 +1,69 @@
+Efficient Approach: The problem can be solved efficiently using Hashing (unordered_map),
+
+    Use hashing to count the frequency of each element of both arrays. Then traverse through the hashtables of the arrays and match the frequencies of the elements.
+
+Follow the below mentioned steps to solve the problem:
+
+    Check if the length of both the arrays are equal or not
+    Create an unordered map and store all elements and frequency of elements of arr1[] in the map.
+    Traverse over arr2[] and check if count of each and every element in arr2[] matches with the count in arr1[]. This can be done by decrementing the frequency while traversing in arr2[].
+
+Below is the implementation of the above approach:
+
+// C++ program to implement the approach
+ 
+#include <bits/stdc++.h>
+using namespace std;
+ 
+// Function to check if both arrays are equal
+bool checkArrays(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 frequencies 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 Code
+int main()
+{
+    int arr1[] = { 1, 2, 3, 4, 5 };
+    int arr2[] = { 4, 3, 1, 5, 2 };
+    int N = sizeof(arr1) / sizeof(int);
+    int M = sizeof(arr2) / sizeof(int);
+ 
+    // Function call
+    if (checkArrays(arr1, arr2, N, M))
+        cout << "Equal";
+    else
+        cout << "Not Equal";
+    return 0;
+}