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Find whether an array is subset of another array using Map
  • Difficulty Level : Basic
  • Last Updated : 27 Jan, 2020

Given two arrays: arr1[0..m-1] and arr2[0..n-1]. Find whether arr2[] is a subset of arr1[] or not. Both the arrays are not in sorted order. It may be assumed that elements in both arrays are distinct.

Examples:

Input: arr1[] = {11, 1, 13, 21, 3, 7},  arr2[] = {11, 3, 7, 1}
Output: arr2[] is a subset of arr1[]

Input: arr1[] = {1, 2, 3, 4, 5, 6},  arr2[] = {1, 2, 4}
Output: arr2[] is a subset of arr1[]

Input: arr1[] = {10, 5, 2, 23, 19},  arr2[] = {19, 5, 3}
Output: arr2[] is not a subset of arr1[]

Simple Approach: A simple approach is to run two nested loops. The outer loop picks all the elements of B[] one by one. The inner loop linearly searches for the element picked by the outer loop in A[]. If all elements are found, then print Yes, else print No. You can check the solution here.

Efficient Approach: Create a map to store the frequency of each distinct number present in A[]. Then we will check if each number of B[] is present in map or not. If present in the map, we will decrement the frequency value for that number by one and check for the next number. If map value for any number becomes zero, we will erase it from the map. If any number of B[] is not found in the map, we will set the flag value and break the loops and print No. Otherwise, we will print Yes.

C++




// C++ program to check if an array is
// subset of another array
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if an array is
// subset of another array
  
int isSubset(int a[], int b[], int m, int n)
{
  
    // map to store the values of array a[]
    map<int, int> mp1;
  
    for (int i = 0; i < m; i++)
        mp1[a[i]]++;
  
    // flag value
    int f = 0;
  
    for (int i = 0; i < n; i++) {
        // if b[i] is not present in map
        // then array b[] can not be a
        // subset of array a[]
  
        if (mp1.find(b[i]) == mp1.end()) {
            f = 1;
            break;
        }
  
        // if if b[i] is present in map
        // decrement by one
        else {
            mp1[b[i]]--;
  
            if (mp1[b[i]] == 0)
                mp1.erase(mp1.find(b[i]));
        }
    }
  
    return f;
}
  
// Driver code
int main()
{
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
  
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
  
    if (!isSubset(arr1, arr2, m, n))
        cout<<"arr2[] is subset of arr1[] ";
    else
        cout<<"arr2[] is not a subset of arr1[]";
  
    return 0;
}


Java




// Java program to check if an array is
// subset of another array
import java.util.*;
  
class GFG
{
  
    // Function to check if an array is
    // subset of another array
    static int isSubset(int a[], int b[], int m, int n) 
    {
  
        // map to store the values of array a[]
        HashMap<Integer, Integer> mp1 = new
                HashMap<Integer, Integer>();
  
        for (int i = 0; i < m; i++)
            if (mp1.containsKey(a[i])) 
            {
                mp1.put(a[i], mp1.get(a[i]) + 1);
            
            else
            {
                mp1.put(a[i], 1);
            }
  
        // flag value
        int f = 0;
  
        for (int i = 0; i < n; i++) 
        {
            // if b[i] is not present in map
            // then array b[] can not be a
            // subset of array a[]
            if (!mp1.containsKey(b[i])) 
            {
                f = 1;
                break;
            }
  
            // if if b[i] is present in map
            // decrement by one
            else 
            {
                mp1.put(b[i], mp1.get(b[i]) - 1);
  
                if (mp1.get(b[i]) == 0)
                    mp1.remove(b[i]);
            }
        }
  
        return f;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 11, 1, 13, 21, 3, 7 };
        int arr2[] = { 11, 3, 7, 1 };
  
        int m = arr1.length;
        int n = arr2.length;
  
        if (isSubset(arr1, arr2, m, n)!=1)
            System.out.print("arr2[] is subset of arr1[] ");
        else
            System.out.print("arr2[] is not a subset of arr1[]");
    }
}
  
// This code is contributed by Rajput-Ji


Python




# Python program to check if an array is 
# subset of another array 
  
# Function to check if an array is 
# subset of another array 
def isSubset(a, b, m, n) :
      
    # map to store the values of array a 
    mp1 = {}
    for i in range(m):
        if a[i] not in mp1:
            mp1[a[i]] = 0
        mp1[a[i]] += 1
      
    # flag value 
    f = 0
    for i in range(n):
          
        # if b[i] is not present in map 
        # then array b can not be a 
        # subset of array a 
        if b[i] not in mp1:
            f = 1
            break
          
        # if if b[i] is present in map 
        # decrement by one 
        else :
            mp1[b[i]] -= 1
              
            if (mp1[b[i]] == 0):
                mp1.pop(b[i])
    return
      
# Driver code 
arr1 = [11, 1, 13, 21, 3, 7 ]
arr2 = [11, 3, 7, 1 ]
  
m = len(arr1) 
n = len(arr2)
  
if (not isSubset(arr1, arr2, m, n)):
    print("arr2[] is subset of arr1[] ")
else:
    print("arr2[] is not a subset of arr1[]")
  
# This code is contributed by Shubhamsingh10


C#




// C# program to check if an array is
// subset of another array
using System;
using System.Collections.Generic;
  
class GFG
{
  
    // Function to check if an array is
    // subset of another array
    static int isSubset(int []a, int []b, int m, int n) 
    {
  
        // map to store the values of array []a
        Dictionary<int, int> mp1 = new
                Dictionary<int, int>();
  
        for (int i = 0; i < m; i++)
            if (mp1.ContainsKey(a[i])) 
            {
                mp1[a[i]] = mp1[a[i]] + 1;
            
            else
            {
                mp1.Add(a[i], 1);
            }
  
        // flag value
        int f = 0;
  
        for (int i = 0; i < n; i++) 
        {
            // if b[i] is not present in map
            // then array []b can not be a
            // subset of array []a
            if (!mp1.ContainsKey(b[i])) 
            {
                f = 1;
                break;
            }
  
            // if if b[i] is present in map
            // decrement by one
            else
            {
                mp1[b[i]] = mp1[b[i]] - 1;
  
                if (mp1[b[i]] == 0)
                    mp1.Remove(b[i]);
            }
        }
  
        return f;
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int []arr1 = {11, 1, 13, 21, 3, 7};
        int []arr2 = {11, 3, 7, 1};
  
        int m = arr1.Length;
        int n = arr2.Length;
  
        if (isSubset(arr1, arr2, m, n) != 1)
            Console.Write("arr2[] is subset of arr1[] ");
        else
            Console.Write("arr2[] is not a subset of arr1[]");
    }
}
  
// This code is contributed by PrinciRaj1992


Output:

arr2[] is subset of arr1[]

Time Complexity: O (n)

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