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

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

Time complexity: 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

Time Complexity: 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 the size of vector A and vector B is not equal return False
    if (B.size() != n)
        return false;

    // Create a hash table to
    // number of instances
    unordered_map<int, 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;
}

// driver code to test above function
int main(){
      // initializing vector A
    vector<int> A = {2, 5, 10, 6, 8, 2, 2};

      // initializing vector B
     vector<int> B = {2, 5, 6, 8, 10, 2, 2};

      // Function call
    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

Time Complexity: O(n), where n is the number of elements in the given vector.
Auxiliary Space: O(n)

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