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Check if two unsorted arrays (with duplicates allowed) have same elements

  • Difficulty Level : Easy
  • Last Updated : 21 May, 2021

Given two unsorted arrays, check whether both arrays have the same set of elements or not. 

Examples: 

Input : A = {2, 5, 6, 8, 10, 2, 2}
        B = {2, 5, 5, 6, 8, 5, 6}
Output : No 

Input : A = {2, 5, 6, 8, 2, 10, 2}
        B = {2, 5, 6, 8, 2, 10, 2}
Output : Yes

Input : A = {2, 5, 8, 6, 10, 2, 2}
        B = {2, 5, 6, 8, 2, 10, 2}
Output : Yes

Method 1 (Simple):
A simple solution to this problem is to check if each element of A is present in B. But this approach will lead to a wrong answer in case of multiple instances of an element is present in B. To overcome this issue, we mark visited instances of B[] using an auxiliary array visited[]. 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if both arrays are same
bool areSameSet(vector<int> A, vector<int> B)
{
    int n = A.size();
    if (B.size() != n)
        return false;
 
    // visited array is used to handle duplicates
    vector<bool> visited(n, false);
 
    // each element of A is matched
    // against each element of B
    for (int i = 0; i < n; i++) {
 
        int j = 0;
        for (j = 0; j < n; j++)
        {
            if (A[i] == B[j] && visited[j] == false)
            {
                visited[j] = true;
                break;           
            }  
        }
 
        // If we could not find A[i] in B[]
        if (j == n)
            return false;  
        
    }
    return true;
}
 
// Driver code
int main()
{
    vector<int> A, B;
    A.push_back(2);
    A.push_back(5);
    A.push_back(10);
    A.push_back(6);
    A.push_back(8);
    A.push_back(2);
    A.push_back(2);
 
    B.push_back(2);
    B.push_back(5);
    B.push_back(6);
    B.push_back(8);
    B.push_back(10);
    B.push_back(2);
    B.push_back(2);
 
    areSameSet(A, B)? cout << "Yes" : cout << "No";
}

Java




// Java implementation of the above approach
import java.util.*;
 
class GFG
{
 
    // Function to check if both arrays are same
    static boolean areSameSet(Vector<Integer> A, Vector<Integer> B)
    {
        int n = A.size();
        if (B.size() != n)
        {
            return false;
        }
 
        // visited array is used to handle duplicates
        Vector<Boolean> visited = new Vector<Boolean>();
        for (int i = 0; i < n; i++)
        {
            visited.add(i, Boolean.FALSE);
        }
         
        // each element of A is matched
        // against each element of B
        for (int i = 0; i < n; i++)
        {
 
            int j = 0;
            for (j = 0; j < n; j++)
            {
                if (A.get(i) == B.get(j) && visited.get(j) == false)
                {
                    visited.add(j, Boolean.TRUE);
                    break;
                }
            }
 
            // If we could not find A[i] in B[]
            if (j == n)
            {
                return false;
            }
 
        }
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        Vector<Integer> A = new Vector<>();
        Vector<Integer> B = new Vector<>();
        A.add(2);
        A.add(5);
        A.add(10);
        A.add(6);
        A.add(8);
        A.add(2);
        A.add(2);
 
        B.add(2);
        B.add(5);
        B.add(6);
        B.add(8);
        B.add(10);
        B.add(2);
        B.add(2);
 
        if (areSameSet(A, B))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 implementation of the above approach
 
# Function to check if both arrays are same
def areSameSet(A, B):
 
    n = len(A)
    if (len(B) != n):
        return False
 
    # visited array is used to handle duplicates
    visited = [False for i in range(n)]
 
    # each element of A is matched
    # against each element of B
    for i in range(n):
 
        j = 0
        for j in range(n):
            if (A[i] == B[j] and
                visited[j] == False):
                visited[j] = True
                break
 
        # If we could not find A[i] in B[]
        if (j == n):
            return False
 
    return True
 
# Driver code
A = []
B = []
A.append(2)
A.append(5)
A.append(10)
A.append(6)
A.append(8)
A.append(2)
A.append(2)
 
B.append(2)
B.append(5)
B.append(6)
B.append(8)
B.append(10)
B.append(2)
B.append(2)
 
if(areSameSet(A, B)):
    print("Yes")
else:
    print("No")
     
# This code is contributed
# by mohit kumar

C#




// C# implementation of the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to check if both arrays are same
    static Boolean areSameSet(List<int> A, List<int> B)
    {
        int n = A.Count;
        if (B.Count != n)
        {
            return false;
        }
 
        // visited array is used to handle duplicates
        List<Boolean> visited = new List<Boolean>();
        for (int i = 0; i < n; i++)
        {
            visited.Insert(i, false);
        }
         
        // each element of A is matched
        // against each element of B
        for (int i = 0; i < n; i++)
        {
 
            int j = 0;
            for (j = 0; j < n; j++)
            {
                if (A[i] == B[j] && visited[j] == false)
                {
                    visited.Insert(j, true);
                    break;
                }
            }
 
            // If we could not find A[i] in B[]
            if (j == n)
            {
                return false;
            }
 
        }
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        List<int> A = new List<int>();
        List<int> B = new List<int>();
        A.Add(2);
        A.Add(5);
        A.Add(10);
        A.Add(6);
        A.Add(8);
        A.Add(2);
        A.Add(2);
 
        B.Add(2);
        B.Add(5);
        B.Add(6);
        B.Add(8);
        B.Add(10);
        B.Add(2);
        B.Add(2);
 
        if (areSameSet(A, B))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript




<script>
// Javascript implementation of the above approach
     
     // Function to check if both arrays are same
    function areSameSet(A,B)
    {
        let n = A.length;
        if (B.length != n)
        {
            return false;
        }
  
        // visited array is used to handle duplicates
        let visited = [];
        for (let i = 0; i < n; i++)
        {
            visited.push(false);
        }
          
        // each element of A is matched
        // against each element of B
        for (let i = 0; i < n; i++)
        {
  
            let j = 0;
            for (j = 0; j < n; j++)
            {
                if (A[i] == B[j] && visited[j] == false)
                {
                    visited[j]=true;
                    break;
                }
            }
  
            // If we could not find A[i] in B[]
            if (j == n)
            {
                return false;
            }
  
        }
        return true;
    }
     
    // Driver code
    let A=[];
    let B=[];
    A.push(2);
        A.push(5);
        A.push(10);
        A.push(6);
        A.push(8);
        A.push(2);
        A.push(2);
  
        B.push(2);
        B.push(5);
        B.push(6);
        B.push(8);
        B.push(10);
        B.push(2);
        B.push(2);
  
        if (areSameSet(A, B))
        {
            document.write("Yes");
        }
        else
        {
            document.write("No");
        }
 
 
// This code is contributed by patel2127
</script>

Output: 

Yes

The time complexity of the above solution in O(n^2).



Method 2 (Sorting):
Sort both the arrays and compare corresponding elements of each array. 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if both arrays are same
bool areSameSet(vector<int> A, vector<int> B)
{
    int n = A.size();
    if (B.size() != n)
        return false;
 
    sort(A.begin(), A.end());
    sort(B.begin(), B.end());
 
    // Compare corresponding elements
    for (int i = 0; i < n; i++)
        if (A[i] != B[i])
            return false;       
     
    return true;
}
 
int main()
{
    vector<int> A, B;
    A.push_back(2);
    A.push_back(5);
    A.push_back(10);
    A.push_back(6);
    A.push_back(8);
    A.push_back(2);
    A.push_back(2);
 
    B.push_back(2);
    B.push_back(5);
    B.push_back(6);
    B.push_back(8);
    B.push_back(10);
    B.push_back(2);
    B.push_back(2);
 
    areSameSet(A, B)? cout << "Yes" : cout << "No";
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
    // Function to check if both arrays are same
    static boolean areSameSet(Vector<Integer> A,
                                Vector<Integer> B)
    {
        int n = A.size();
        if (B.size() != n)
        {
            return false;
        }
 
        Collections.sort(A);
        Collections.sort(B);
 
        // Compare corresponding elements
        for (int i = 0; i < n; i++)
        {
            if (A.get(i) != B.get(i))
            {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        Vector<Integer> A = new Vector<>();
        Vector<Integer> B = new Vector<>();
        A.add(2);
        A.add(5);
        A.add(10);
        A.add(6);
        A.add(8);
        A.add(2);
        A.add(2);
 
        B.add(2);
        B.add(5);
        B.add(6);
        B.add(8);
        B.add(10);
        B.add(2);
        B.add(2);
 
        if (areSameSet(A, B))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
 
# Function to check if
# both arrays are same
def areSameSet(A, B):
 
    n = len(A)
    if (len(B) != n):
        return False
 
    A.sort()
    B.sort()
 
    # Compare corresponding
    # elements
    for i in range (n):
        if (A[i] != B[i]):
            return False     
     
    return True
 
# Driver code
if __name__ == "__main__":
 
    A = []
    B = []
    A.append(2)
    A.append(5)
    A.append(10)
    A.append(6)
    A.append(8)
    A.append(2)
    A.append(2)
 
    B.append(2)
    B.append(5)
    B.append(6)
    B.append(8)
    B.append(10)
    B.append(2)
    B.append(2)
 
    if areSameSet(A, B):
         print ("Yes")
    else:
         print ("No")
 
# This code is contributed by Chitranayal

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
    // Function to check if both arrays are same
    static Boolean areSameSet(List<int> A,
                                List<int> B)
    {
        int n = A.Count;
        if (B.Count!= n)
        {
            return false;
        }
 
        A.Sort();
        B.Sort();
 
        // Compare corresponding elements
        for (int i = 0; i < n; i++)
        {
            if (A[i] != B[i])
            {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        List<int> A = new List<int>();
        List<int> B = new List<int>();
        A.Add(2);
        A.Add(5);
        A.Add(10);
        A.Add(6);
        A.Add(8);
        A.Add(2);
        A.Add(2);
 
        B.Add(2);
        B.Add(5);
        B.Add(6);
        B.Add(8);
        B.Add(10);
        B.Add(2);
        B.Add(2);
 
        if (areSameSet(A, B))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to check if both arrays are same
function areSameSet(A, B)
{
    let n = A.length;
    if (B.length != n)
    {
        return false;
    }
 
    A.sort(function(a, b){return a - b;});
    B.sort(function(a, b){return a - b;});
 
    // Compare corresponding elements
    for(let i = 0; i < n; i++)
    {
        if (A[i] != B[i])
        {
            return false;
        }
    }
    return true;
}
 
// Driver code
let A = [];
let B = [];
 
A.push(2);
A.push(5);
A.push(10);
A.push(6);
A.push(8);
A.push(2);
A.push(2);
 
B.push(2);
B.push(5);
B.push(6);
B.push(8);
B.push(10);
B.push(2);
B.push(2);
 
if (areSameSet(A, B))
{
    document.write("Yes");
}
else
{
    document.write("No");
}
 
// This code is contributed by unknown2108
 
</script>

Output: 

Yes

The Time Complexity of the above solution is O(n*log(n)).

Method 3 (Hashing):
We can decrease the time complexity of the above problem by using a Hash table. First, we iterate through A and mark the number of instances of each element of A in a Hash Table. Then we iterate through B and decrease the corresponding value in the hash table. If in the end if all the entries of the hash table are zero, the answer will be “Yes” else “No”. 

C++




// C++ program to implement Naive approach
// to remove duplicates
#include <bits/stdc++.h>
using namespace std;
 
bool areSameSet(vector<int> A, vector<int> B)
{
    int n = A.size();
    if (B.size() != n)
        return false;
 
    // Create a hash table to
    // number of instances
    unordered_map<int> m;
 
    // for each element of A
    // increase it's instance by 1.
    for (int i = 0; i < n; i++)
        m[A[i]]++;
     
    // for each element of B
    // decrease it's instance by 1.
    for (int i = 0; i < n; i++)
        m[B[i]]--;
     
    // Iterate through map and check if
    // any entry is non-zero
    for (auto i : m)
        if (i.second != 0)
            return false;
     
    return true;
}
 
int main()
{
    vector<int> A, B;
    A.push_back(2);
    A.push_back(5);
    A.push_back(10);
    A.push_back(6);
    A.push_back(8);
    A.push_back(2);
    A.push_back(2);
 
    B.push_back(2);
    B.push_back(5);
    B.push_back(6);
    B.push_back(8);
    B.push_back(10);
    B.push_back(2);
    B.push_back(2);
 
    areSameSet(A, B)? cout << "Yes" : cout << "No";
}

Java




// Java program to implement Naive approach
// to remove duplicates
import java.util.HashMap;
 
class GFG
{
static boolean areSameSet(int[] A, int[] B)
{
    int n = A.length;
 
    if (B.length != n)
        return false;
 
    // Create a hash table to
    // number of instances
    HashMap<Integer,
            Integer> m = new HashMap<>();
 
    // for each element of A
    // increase it's instance by 1.
    for (int i = 0; i < n; i++)
        m.put(A[i], m.get(A[i]) == null ? 1 :
                    m.get(A[i]) + 1);
 
    // for each element of B
    // decrease it's instance by 1.
    for (int i = 0; i < n; i++)
        m.put(B[i], m.get(B[i]) - 1);
 
    // Iterate through map and check if
    // any entry is non-zero
    for (HashMap.Entry<Integer,
                       Integer> entry : m.entrySet())
        if (entry.getValue() != 0)
            return false;
    return true;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] A = { 2, 5, 10, 6, 8, 2, 2 };
    int[] B = { 2, 5, 6, 8, 10, 2, 2 };
 
    if (areSameSet(A, B))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by
// sanjeev2552

Python3




# Python3 program to implement Naive
# approach to remove duplicates
def areSameSet(A, B):
     
    n = len(A)
     
    if (len(B) != n):
        return False
     
    # Create a hash table to
    # number of instances
    m = {}
     
    # For each element of A
    # increase it's instance by 1.
    for i in range(n):
        if A[i] not in m:
            m[A[i]] = 1
        else:
            m[A[i]] += 1
             
    # For each element of B
    # decrease it's instance by 1.
    for i in range(n):
        if B[i] in m:
            m[B[i]] -= 1
 
    # Iterate through map and check if
    # any entry is non-zero
    for i in m:
        if (m[i] != 0):
            return False
             
    return True
 
# Driver Code
A = []
B = []
 
A.append(2)
A.append(5)
A.append(10)
A.append(6)
A.append(8)
A.append(2)
A.append(2)
 
B.append(2)
B.append(5)
B.append(6)
B.append(8)
B.append(10)
B.append(2)
B.append(2)
 
if (areSameSet(A, B)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by avanitrachhadiya2155

C#




// C# program to implement Naive approach
// to remove duplicates
using System;
using System.Collections.Generic;
 
class GFG
{
     
static bool areSameSet(int[] A, int[] B)
{
    int n = A.Length;
 
    if (B.Length != n)
        return false;
 
    // Create a hash table to
    // number of instances
    Dictionary<int,int> m = new Dictionary<int,int>();
 
    // for each element of A
    // increase it's instance by 1.
    for (int i = 0; i < n; i++)
        if(m.ContainsKey(A[i]))
                m[A[i]] = m[A[i]] + 1;
        else
            m.Add(A[i], 1);
 
 
    // for each element of B
    // decrease it's instance by 1.
    for (int i = 0; i < n; i++)
        if(m.ContainsKey(B[i]))
                m[B[i]] = m[B[i]] - 1;
 
 
    // Iterate through map and check if
    // any entry is non-zero
    foreach(KeyValuePair<int, int> entry in m)
        if (entry.Value != 0)
            return false;
    return true;
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] A = { 2, 5, 10, 6, 8, 2, 2 };
    int[] B = { 2, 5, 6, 8, 10, 2, 2 };
 
    if (areSameSet(A, B))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript program to implement Naive approach
// to remove duplicates
     
    function areSameSet(A,B)
    {
        let n = A.length;
  
    if (B.length != n)
        return false;
  
    // Create a hash table to
    // number of instances
    let m = new Map();
  
    // for each element of A
    // increase it's instance by 1.
    for (let i = 0; i < n; i++)
        m.set(A[i], m.get(A[i]) == null ? 1 :
                    m.get(A[i]) + 1);
  
    // for each element of B
    // decrease it's instance by 1.
    for (let i = 0; i < n; i++)
        m.set(B[i], m.get(B[i]) - 1);
  
    // Iterate through map and check if
    // any entry is non-zero
    for (let [key, value] of m.entries())
        if (value != 0)
            return false;
    return true;
    }
     
    // Driver Code
    let A=[2, 5, 10, 6, 8, 2, 2 ];
    let B=[2, 5, 6, 8, 10, 2, 2];
    if (areSameSet(A, B))
        document.write("Yes");
    else
        document.write("No");
     
 
// This code is contributed by rag2127
 
</script>

Output: 

Yes

The time complexity of the above method is O(n).
This article is contributed by Raghav Sharma. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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