Multiset Equivalence Problem

Unlike a set, a multiset may contain multiple occurrences of same number. The multiset equivalence problem states to check if two given multisets are equal or not. For example let A = {1, 2, 3} and B = {1, 1, 2, 3}. Here A is set but B is not (1 occurs twice in B), whereas A and B are both multisets. More formally, “Are the sets of pairs defined as  \(A' = \{ (a, frequency(a)) | a \in \mathbf{A} \}\) equal for the two given multisets?”

Given two multisets A and B, write a program to check if the two multisets are equal.

Note: Elements in the multisets can be of order 109

Examples:

Input : A = {1, 1, 3, 4},  
        B = {1, 1, 3, 4}
Output : Yes

Input : A = {1, 3},  
        B = {1, 1}
Output : No

Since the elements are as large as 10^9 we cannot use direct index table.



One solution is to sort both multisets and compare them one by one.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to check if two given multisets
// are equivalent
#include <bits/stdc++.h>
using namespace std;
  
bool areSame(vector<int>& a, vector<int>& b)
{
    // sort the elements of both multisets
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
  
    // Return true if both multisets are same.
    return (a == b);
}
  
int main()
{
    vector<int> a({ 7, 7, 5 }), b({ 7, 5, 5 });
    if (areSame(a, b))
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to check if two given multisets 
// are equivalent 
import java.util.*;
class GFG {
  
  
static boolean areSame(Vector<Integer>a, Vector<Integer>b) 
    // sort the elements of both multisets 
    Collections.sort(a); 
    Collections.sort(b); 
  
    // Return true if both multisets are same. 
    return (a == b); 
 public static void main(String[] args) {
       Vector<Integer> a = new Vector<Integer>(Arrays.asList( 7, 7, 5 ));
       Vector<Integer> b = new Vector<Integer>(Arrays.asList( 7, 5, 5));
    if (areSame(a, b)) 
        System.out.print("Yes\n"); 
    else
        System.out.print("No\n");
    }
}
// This code is contributed by PrinciRaj1992

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to check if 
# two given multisets are equivalent
  
def areSame(a, b):
      
    # sort the elements of both multisets
    a.sort();
    b.sort();
  
    # Return true if both multisets are same.
    return (a == b);
  
# Driver Code
a = [ 7, 7, 5 ];
b = [ 7, 5, 5 ];
if (areSame(a, b)):
    print("Yes");
else:
    print("No");
  
# This code is contributed by Princi Singh

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to check if two given multisets 
// are equivalent 
using System;
using System.Collections.Generic;
  
class GFG
{
  
static bool areSame(List<int>a, List<int>b) 
    // sort the elements of both multisets 
    a.Sort();
    b.Sort();
  
    // Return true if both multisets are same. 
    return (a == b); 
  
// Driver code
public static void Main()
{
    List<int> a = new List<int> { 7, 7, 5 };
    List<int> b = new List<int> { 7, 5, 5 };
    if (areSame(a, b)) 
        Console.WriteLine("Yes\n"); 
    else
        Console.WriteLine("No\n");
}
}
  
// This code is contributed by Rajput-Ji

chevron_right


Output:

No

A better solution is to use hashing. We create two empty hash tables (implemented using unordered_map in C++). We first insert all items of first multimap in first table and all items of second multiset in second table. Now we check if both hash tables contain same items and frequencies or not.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to check if two given multisets
// are equivalent
#include <bits/stdc++.h>
using namespace std;
  
bool areSame(vector<int>& a, vector<int>& b)
{
    if (a.size() != b.size())
        return false;
  
    // Create two unordered maps m1 and m2
    // and insert values of both vectors.
    unordered_map<int, int> m1, m2;
    for (int i = 0; i < a.size(); i++) {
        m1[a[i]]++;
        m2[b[i]]++;
    }
  
    // Now we check if both unordered_maps
    // are same of not.
    for (auto x : m1) {
        if (m2.find(x.first) == m2.end() || 
            m2[x.first] != x.second)
            return false;
    }
  
    return true;
}
  
// Driver code
int main()
{
    vector<int> a({ 7, 7, 5 }), b({ 7, 7, 5 });
    if (areSame(a, b))
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to check if two given multisets
// are equivalent
import java.util.*;
  
class GFG 
{
static boolean areSame(int []a, int []b)
{
    if (a.length != b.length)
        return false;
  
    // Create two unordered maps m1 and m2
    // and insert values of both vectors.
    HashMap<Integer, Integer> m1, m2;
    m1 = new HashMap<Integer, Integer>();
    m2 = new HashMap<Integer, Integer>();
    for (int i = 0; i < a.length; i++)
    {
        if(m1.containsKey(a[i]))
        {
            m1.put(a[i], m1.get(a[i]) + 1);
        }
        else
        {
            m1.put(a[i], 1);
        }
        if(m2.containsKey(b[i]))
        {
            m2.put(b[i], m2.get(b[i]) + 1);
        }
        else
        {
            m2.put(b[i], 1);
        }
    }
  
    // Now we check if both unordered_maps
    // are same of not.
    for (Map.Entry<Integer, Integer> x : m1.entrySet())
    {
        if (!m2.containsKey(x.getKey()) || 
             m2.get(x.getKey()) != x.getValue())
            return false;
    }
    return true;
}
  
// Driver code
public static void main(String args[]) 
{
    int []a = { 7, 7, 5 };
    int []b = { 7, 7, 5 };
    if (areSame(a, b))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to check if two given multisets
// are equivalent 
using System;
using System.Collections.Generic;
  
class GFG 
{
static bool areSame(int []a, int []b)
{
    if (a.Length != b.Length)
        return false;
  
    // Create two unordered maps m1 and m2
    // and insert values of both vectors.
    Dictionary<int, int> m1, m2;
    m1 = new Dictionary<int, int>();
    m2 = new Dictionary<int, int>();
    for (int i = 0; i < a.Length; i++)
    {
        if(m1.ContainsKey(a[i]))
        {
            m1[a[i]] = m1[a[i]] + 1;
        }
        else
        {
            m1.Add(a[i], 1);
        }
        if(m2.ContainsKey(b[i]))
        {
            m2[b[i]] = m2[b[i]] + 1;
        }
        else
        {
            m2.Add(b[i], 1);
        }
    }
  
    // Now we check if both unordered_maps
    // are same of not.
    foreach(KeyValuePair<int, int> x in m1)
    {
        if (!m2.ContainsKey(x.Key) || 
             m2[x.Key] != x.Value)
            return false;
    }
    return true;
}
  
// Driver code
public static void Main(String []args) 
{
    int []a = { 7, 7, 5 };
    int []b = { 7, 7, 5 };
    if (areSame(a, b))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

Yes

Time complexity : O(n) under the assumption that unordered_map find() and insert() operations work in O(1) time.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.