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Check if given two Arrays are equal (using Map)

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  • Difficulty Level : Easy
  • Last Updated : 08 Sep, 2022
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Given two arrays, A[] and B[], the task is to check if they are equal or not. Arrays are considered equal if any permutation of array B equates to array A.

Examples:

Input: A[] = [2, 4, 5, 7, 5, 6] and B[] = [4, 2, 5, 5, 6, 7]
Output: Yes
Explanation: All the elements in array A are present in array B and same number of times.

Input: A[] = [2, 5, 8, 9, 78] and B[] = [5, 2, 7, 78, 8]
Output: No
Explanation: In array A there is a 9 and in array B there is a 7

Approach: The problem can be solved using hashmap based on the below idea:

Two arrays will be equal only if the frequency of the respective elements in both arrays are equal.

Follow the steps to solve the problem:

  • If the sizes of both arrays are not equal, return NO.
  • Maintain a hashmap and count the frequency of both the array elements
  • If for any element the frequency is not the same, then return NO
  • Otherwise, return YES

Below is the implementation of the above approach.

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
bool isEqual(vector<int> a, vector<int> b, int n, int m)
{
    // If size is not same return false
    if (n != m) {
        return 0;
    }
 
    // Create 2 unordered maps to store
    // the frequency
    unordered_map<int, int> mp1, mp2;
    for (int i : a) {
        mp1[i]++;
    }
    for (int i : b) {
        mp2[i]++;
    }
 
    // Compare the frequency
    for (auto i = mp1.begin(); i != mp1.end(); i++) {
 
        // If frequency not same return false
        if (mp2[i->first] != i->second) {
            return 0;
        }
    }
    return 1;
}
 
// Drivers code
int main()
{
    vector<int> a = { 2, 4, 5, 7, 5, 6 },
                b = { 4, 2, 5, 5, 6, 7 };
    int n = a.size(), m = b.size();
    bool flag = isEqual(a, b, n, m);
 
    // Return 1 if equal
    if (flag) {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
    return 0;
}

Java




// Java code to implement the approach
import java.io.*;
import java.util.*;
 
class GFG {
    public static boolean isEqual(int a[], int b[], int n,
                                  int m)
    {
        // If size is not same return false
        if (n != m) {
            return false;
        }
 
        // Create 2 unordered maps to store
        // the frequency
        HashMap<Integer, Integer> mp1
            = new HashMap<Integer, Integer>();
        HashMap<Integer, Integer> mp2
            = new HashMap<Integer, Integer>();
        for (int i : a) {
            if (mp1.get(i) != null)
                mp1.put(i, mp1.get(i) + 1);
            else
                mp1.put(i, 1);
        }
        for (int i : b) {
            if (mp2.get(i) != null)
                mp2.put(i, mp2.get(i) + 1);
            else
                mp2.put(i, 1);
        }
 
        // Compare the frequency
        for (Map.Entry<Integer, Integer> i :
             mp1.entrySet()) {
            Integer key = i.getKey();
            // If frequency not same return false
            if (mp2.get(key) != i.getValue()) {
                return false;
            }
        }
        return true;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int a[] = { 2, 4, 5, 7, 5, 6 };
        int b[] = { 4, 2, 5, 5, 6, 7 };
        int n = a.length, m = b.length;
        boolean flag = isEqual(a, b, n, m);
 
        // Return 1 if equal
        if (flag == true) {
            System.out.print("YES\n");
        }
        else {
            System.out.print("NO\n");
        }
    }
}
 
// This code is contributed by Rohit Pradhan

Python3




# Python3 code to implement the approach
 
def isEqual(a, b, n, m) :
 
    # If size is not same return false
    if (n != m) :
        return 0
 
    # Create 2 unordered maps to store
    # the frequency
    mp1 = dict.fromkeys(a, 0);
    mp2 = dict.fromkeys(b, 0);
     
    for i in a :
        if i in mp1 :
            mp1[i] += 1;
        else:
            mp1[i] = 0
         
    for i in b :
        if i in mp2 :
            mp2[i] += 1;
        else :
            mp2[i] = 0
 
    # Compare the frequency
    for i in mp1 :
 
        # If frequency not same return false
        if (mp2[i] != mp1[i]) :
            return 0;
 
    return 1;
 
# Drivers code
if __name__ == "__main__" :
 
    a = [ 2, 4, 5, 7, 5, 6 ];
    b = [ 4, 2, 5, 5, 6, 7 ];
    n = len(a);
    m = len(b);
    flag = isEqual(a, b, n, m);
 
    # Return 1 if equal
    if (flag) :
        print("YES");
 
    else :
        print("NO");
   
  # This code is contributed by AnkThon

C#




// C# code to implement the approach
using System;
using System.Collections.Generic;
 
public class GFG {
  public static bool isEqual(int []a, int []b, int n,
                             int m)
  {
    // If size is not same return false
    if (n != m) {
      return false;
    }
 
    // Create 2 unordered maps to store
    // the frequency
    Dictionary<int, int> mp1
      = new Dictionary<int, int>();
    Dictionary<int, int> mp2
      = new Dictionary<int, int>();
    foreach (int i in a) {
      if (mp1.ContainsKey(i))
        mp1[i] = mp1[i] + 1;
      else
        mp1.Add(i, 1);
    }
    foreach (int i in b) {
      if (mp2.ContainsKey(i))
        mp2[i] = mp2[i] + 1;
      else
        mp2.Add(i, 1);
    }
 
    // Compare the frequency
    foreach (KeyValuePair<int, int> i in
             mp1) {
      int key = i.Key;
       
      // If frequency not same return false
      if (mp2[key] != i.Value) {
        return false;
      }
    }
    return true;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []a = { 2, 4, 5, 7, 5, 6 };
    int []b = { 4, 2, 5, 5, 6, 7 };
    int n = a.Length, m = b.Length;
    bool flag = isEqual(a, b, n, m);
 
    // Return 1 if equal
    if (flag == true) {
      Console.Write("YES\n");
    }
    else {
      Console.Write("NO\n");
    }
  }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
// Javascript code to implement the approach
function isEqual(a, b, n, m)
{
 
    // If size is not same return false
    if (n != m) {
        return false;
    }
 
    // Create 2 unordered maps to store
    // the frequency
    let mp1 = new Map();
    let mp2 = new Map();
    for (let i of a) {
        if (mp1.get(i) != null)
            mp1.set(i, mp1.get(i) + 1);
        else
            mp1.set(i, 1);
    }
    for (let i of b) {
        if (mp2.get(i) != null)
            mp2.set(i, mp2.get(i) + 1);
        else
            mp2.set(i, 1);
    }
 
    // Compare the frequency
    for (let i of mp1.keys()) {
        // If frequency not same return false
        if (mp2.get(i) != mp1.get(i)) {
            return false;
        }
    }
    return true;
}
 
// Driver Code
 
let a = [2, 4, 5, 7, 5, 6];
let b = [4, 2, 5, 5, 6, 7];
let n = a.length, m = b.length;
let flag = isEqual(a, b, n, m);
 
// Return 1 if equal
if (flag == true) {
    document.write("YES<br>");
}
else {
    document.write("NO<br>");
}
 
// This code is contributed by Saurabh Jaiswal
</script>

Output

YES

Time complexity: O(N) in average case and O(N2) in worst case.

Auxiliary Space: O(N)


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