Union and Intersection of two Linked List using Hashing

Check if an array is subset of another array

Last Updated : 31 Dec, 2024

Try it on GfG Practice

Given two arrays a[] and b[] of size m and n respectively, the task is to determine whether b[] is a subset of a[]. Both arrays are not sorted, and elements are distinct.

Examples:

Input: a[] = [11, 1, 13, 21, 3, 7], b[] = [11, 3, 7, 1]
Output: true

Input: a[]= [1, 2, 3, 4, 5, 6], b = [1, 2, 4]
Output: true

Input: a[] = [10, 5, 2, 23, 19], b = [19, 5, 3]
Output: false

Table of Content

[Naive approach] Using Nested Loops – O(m*n) Time and O(1) Space
[Better Approach] Using Sorting and Two Pointer
[Expected Approach] Using Hashing- O(m + n) Time and O(m) Space

[Naive approach] Using Nested Loops – O(m*n) Time and O(1) Space

The very basic approach is to use two nested loops: the outer loop picks each element from b[], and the inner loop searches for this element in a[] and check for all elements in b[].

C++

#include<bits/stdc++.h>
using namespace std;

bool isSubset(vector<int> & a, vector<int> & b) {
  
    // Iterate over each element in the second array
  int m=a.size(),n=b.size();
    for (int i = 0; i < n; i++) {
        bool found = false;
      
        // Check if the element exists in the first array
        for (int j = 0; j < m; j++) {
            if (b[i] == a[j]) {
                found = true;
                break;
            }
        }
      
        // If any element is not found, return false
        if (!found) return false;
    }
  
    // If all elements are found, return true
    return true;
}

int main() {
    vector<int> a = {11, 1, 13, 21, 3, 7};
    vector<int> b = {11, 3, 7, 1};
  
    if (isSubset(a, b)) {
        cout << "true" << endl;
    } else {
        cout << "false" << endl;
    }

    return 0;
}

C

#include <stdio.h>
#include <stdbool.h>

bool isSubset(int a[], int m, int b[], int n) {
    // Iterate over each element in the second array
    for (int i = 0; i < n; i++) {
        bool found = false;
        
        // Check if the element exists in the first array
        for (int j = 0; j < m; j++) {
            if (b[i] == a[j]) {
                found = true;
                break;
            }
        }
        
        // If any element is not found, return false
        if (!found) return false;
    }
    
    // If all elements are found, return true
    return true;
}

int main() {
    int a[] = {11, 1, 13, 21, 3, 7};
    int b[] = {11, 3, 7, 1};
    int m = sizeof(a) / sizeof(a[0]);
    int n = sizeof(b) / sizeof(b[0]);
    
    if (isSubset(a, m, b, n)) {
        printf("true\n");
    } else {
        printf("false\n");
    }
    return 0;
}

Java

 class GfG {

     static boolean isSubset(int[] a, int[] b) {
        // Iterate over each element in the second array
        int m = a.length;
        int n = b.length;
        for (int i = 0; i < n; i++) {
            boolean found = false;
            // Check if the element exists in the first array
            for (int j = 0; j < m; j++) {
                if (b[i] == a[j]) {
                    found = true;
                    break;
                }
            }
            // If any element is not found, return false
            if (!found) return false;
        }
        // If all elements are found, return true
        return true;
    }

    public static void main(String[] args) {
        int[] a = {11, 1, 13, 21, 3, 7};
        int[] b = {11, 3, 7, 1};

        if (isSubset(a, b)) {
            System.out.println("true");
        } else {
            System.out.println("false");
        }
    }
}

Python

def isSubset(a, b):
    m = len(a)
    n = len(b)
    # Iterate over each element in the second array
    for i in range(n):
        found = False

        # Check if the element exists in the first array
        for j in range(m):
            if b[i] == a[j]:
                found = True
                break

        # If any element is not found, return false
        if not found:
            return False

    # If all elements are found, return true
    return True


if __name__ == "__main__":
    a = [11, 1, 13, 21, 3, 7]
    b = [11, 3, 7, 1]

    if isSubset(a, b):
        print("true")
    else:
        print("false")

C#

using System;

class GfG {
    static bool IsSubset(int[] a, int[] b) {
        int m = a.Length;
        int n = b.Length;
        // Iterate over each element in the second array
        for (int i = 0; i < n; i++) {
            bool found = false;
            
            // Check if the element exists in the first array
            for (int j = 0; j < m; j++) {
                if (b[i] == a[j]) {
                    found = true;
                    break;
                }
            }
            
            // If any element is not found, return false
            if (!found) return false;
        }
        
        // If all elements are found, return true
        return true;
    }

    static void Main() {
        int[] a = {11, 1, 13, 21, 3, 7};
        int[] b = {11, 3, 7, 1};
        
        if (IsSubset(a, b)) {
            Console.WriteLine("true");
        } else {
            Console.WriteLine("false");
        }
    }
}

JavaScript

function isSubset(a, b) {

  const m = a.length;
  const n = b.length;
  // Iterate over each element in the second array
  for (let i = 0; i < n; i++) {
    let found = false;
    
    // Check if the element exists in the first array
    for (let j = 0; j < m; j++) {
      if (b[i] === a[j]) {
        found = true;
        break;
      }
    }
    
    // If any element is not found, return false
    if (!found) return false;
  }
  
  // If all elements are found, return true
  return true;
}

//Driver Code
const a = [11, 1, 13, 21, 3, 7];
const b = [11, 3, 7, 1];

if (isSubset(a, b)) {
  console.log("true");
} else {
  console.log("false");
}

Output

true

Time Complexity: O(m*n)
Auxiliary Space: O(1)

[Better Approach] Using Sorting and Two Pointer

Sort both arrays and use two pointers to traverse them. Move the pointer in b[] if a[]’s element is smaller. If a[]’s element is larger, b’s element isn’t in a[], so return false.

C++

#include <bits/stdc++.h>
using namespace std;

bool isSubset(vector<int>& a, vector<int>& b) {

    // Sort both arrays
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());

    int i = 0, j = 0;

    // Traverse both arrays using two pointers
    while (i < a.size() && j < b.size()) {
        if (a[i] < b[j]) {
            i++;
        }
        else if (a[i] == b[j]) {
            i++;
            j++;
        }
        else {
            // If element in b is not found in a
            return false;
        }
    }

    // If we have traversed all elements in b, it is a
    // subset
    return (j == b.size());
}

int main() {
    vector<int> a = { 11, 1, 13, 21, 3, 7 };
    vector<int> b = { 11, 3, 7, 1 };

    if (isSubset(a, b)) {
        cout << "true" << endl;
    }
    else {
        cout << "false" << endl;
    }

    return 0;
}

C

#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

bool isSubset(int a[], int aSize, int b[], int bSize) {
    qsort(a, aSize, sizeof(int), compare);
    qsort(b, bSize, sizeof(int), compare);

    int i = 0, j = 0;

    // Traverse both arrays using two pointers
    while (i < aSize && j < bSize) {
        if (a[i] < b[j]) {
            // Element in a is smaller, move to the next element in a
            i++;
        } else if (a[i] == b[j]) {
            // Element found in both arrays, move to the next element in both arrays
            i++;
            j++;
        } else {
            // Element in b not found in a, not a subset
            return false;
        }
    }

    // If we have traversed all elements in b, it is a subset
    return j == bSize;
}

int main() {
    int a[] = {11, 1, 13, 21, 3, 7};
    int b[] = {11, 3, 7, 1};
    int aSize = sizeof(a) / sizeof(a[0]);
    int bSize = sizeof(b) / sizeof(b[0]);

    if (isSubset(a, aSize, b, bSize)) {
        printf("true\n");
    } else {
        printf("false\n");
    }
    return 0;
}

Java

import java.util.Arrays;

class GfG {
     static boolean isSubset(int[] a, int[] b) {
        Arrays.sort(a);
        Arrays.sort(b);

        int i = 0;
        int j = 0;

        // Traverse both arrays using two pointers
        while (i < a.length && j < b.length) {
            if (a[i] < b[j]) {
                // Element in a is smaller, move to the next element in a
                i++;
            } else if (a[i] == b[j]) {
                // Element found in both arrays, move to the next element in both arrays
                i++;
                j++;
            } else {
                // Element in b not found in a, not a subset
                return false;
            }
        }

        // If we have traversed all elements in b, it is a subset
        return j == b.length;
    }

    public static void main(String[] args) {
        int[] a = {11, 1, 13, 21, 3, 7};
        int[] b = {11, 3, 7, 1};

        if (isSubset(a, b)) {
            System.out.println("true");
        } else {
            System.out.println("false");
        }
    }
}

Python

def isSubset(a, b):

    # Sort both arrays in ascending order
    a.sort()
    b.sort()

    i = 0
    j = 0

    # Traverse both arrays using two pointers
    while i < len(a) and j < len(b):
        if a[i] < b[j]:
            # Element in arr1 is smaller, move to the next element in a
            i += 1
        elif a[i] == b[j]:
            # Element found in both arrays, move to the next element in both arrays
            i += 1
            j += 1
        else:
            # Element in arr2 not found in a, not a subset
            return False

    # If we have traversed all elements in b, it is a subset
    return j == len(b)

if __name__ == "__main__":
  a = [11, 1, 13, 21, 3, 7]
  b = [11, 3, 7, 1]

  if isSubset(a, b):
      print("true")
  else:
      print("false")

C#

using System;
using System.Collections.Generic;
using System.Linq;

class GfG {
    public static bool IsSubset(List<int> a, List<int> b) {
        // Sort both arrays in ascending order
        a.Sort();
        b.Sort();

        int i = 0;
        int j = 0;

        // Traverse both arrays using two pointers
        while (i < a.Count && j < b.Count) {
            if (a[i] < b[j]) {
                // Element in arr1 is smaller, move to the
                // next element in arr1
                i++;
            }
            else if (a[i] == b[j]) {
                // Element found in both arrays, move to the
                // next element in both arrays
                i++;
                j++;
            }
            else {
                // Element in arr2 not found in arr1, not a
                // subset
                return false;
            }
        }

        // If we have traversed all elements in arr2, it is
        // a subset
        return j == b.Count;
    }

    public static void Main(string[] args) {
        List<int> a
            = new List<int>() { 11, 1, 13, 21, 3, 7 };
        List<int> b = new List<int>() { 11, 3, 7, 1 };

        if (IsSubset(a, b)) {
            Console.WriteLine("true");
        }
        else {
            Console.WriteLine("false");
        }
    }
}

JavaScript

function isSubset(a, b) {

    a.sort((a, b) => a - b);
    b.sort((a, b) => a - b);

    let i = 0;
    let j = 0;

    // Traverse both arrays using two pointers
    while (i < a.length && j < b.length) {
        if (a[i] < b[j]) {
            // Element in a is smaller, move to the next element in a
            i++;
        } else if (a[i] === b[j]) {
            // Element found in both arrays, move to the next element in both arrays
            i++;
            j++;
        } else {
            // Element in b not found in a, not a subset
            return false;
        }
    }

    // If we have traversed all elements in b, it is a subset
    return j === b.length;
}

//Driver code
const a = [11, 1, 13, 21, 3, 7];
const b = [11, 3, 7, 1];

if (isSubset(a, b)) {
    console.log("true");
} else {
    console.log("false");
}

Output

true

Time Complexity: O(m log m + n log n)
Auxiliary Space: O(1)

[Expected Approach] Using Hashing- O(m + n) Time and O(m) Space

We can use a hash set to store elements of a[], this will help us in constant time complexity searching. We first insert all elements of a[] into a hash set. Then, for each element in b[], we check if it exists in the hash set.

C++

#include <bits/stdc++.h>
using namespace std;

bool isSubset( vector<int>& a,  vector<int>& b) {

    // Create a hash set and insert all elements of a
    unordered_set<int> hashSet(a.begin(), a.end());
    
    // Check each element of b in the hash set
    for (int num : b) {
        if (hashSet.find(num) == hashSet.end()) {
            return false;
        }
    }
    
    // If all elements of b are found in the hash set
    return true;
}

int main() {
    vector<int> a = {1, 2, 3, 4, 5, 6, 7, 8};
    vector<int> b = {1, 2, 3, 1};
    
    if (isSubset(a, b)) {
        cout << "true" << endl;
    } else {
        cout << "false" << endl;
    }
    
    return 0;
}

Java

import java.util.HashSet;
import java.util.Set;

class GfG {

    static boolean isSubset(int[] a,int[] b) {
        // Create a hash set and insert all elements of a
        Set<Integer> hashSet = new HashSet<>();
        for (int num : a) {
            hashSet.add(num);
        }

        // Check each element of b in the hash set
        for (int num : b) {
            if (!hashSet.contains(num)) {
                return false;
            }
        }

        // If all elements of b are found in the hash set
        return true;
    }

    public static void main(String[] args){
        int[] a = { 11, 1, 13, 21, 3, 7 };
        int[] b = { 11, 3, 7, 1 };

        if (isSubset(a, b)) {
            System.out.println("true");
        }
        else {
            System.out.println("false");
        }
    }
}

Python

def isSubset(a, b):

    # Create a hash set and insert all elements of arr1
    hash_set = set(a)

    # Check each element of arr2 in the hash set
    for num in b:
        if num not in hash_set:
            return False

    # If all elements of arr2 are found in the hash set
    return True


if __name__ == "__main__":
  a = [11, 1, 13, 21, 3, 7]
  b = [11, 3, 7, 1]

  if isSubset(a, b):
      print("true")
  else:
      print("false")

C#

using System;
using System.Collections.Generic;

class GfG{
    static bool isSubset(List<int> a, List<int> b)  {
        // Create a hash set and insert all elements of arr1
        HashSet<int> hashSet = new HashSet<int>(a);

        // Check each element of arr2 in the hash set
        foreach (int num in b){
            if (!hashSet.Contains(num)){
                return false;
            }
        }

        // If all elements of arr2 are found in the hash set
        return true;
    }

    public static void Main(string[] args) {
        List<int> a = new List<int>() { 11, 1, 13, 21, 3, 7 };
        List<int> b = new List<int>() { 11, 3, 7, 1 };

        if (isSubset(a, b))  {
            Console.WriteLine("true");
        }
        else{
            Console.WriteLine("false");
        }
    }
}

JavaScript

function isSubset(a, b) {

    // Create a hash set and insert all elements of a
    const hashSet = new Set(a);

    // Check each element of b in the hash set
    for (const num of b) {
        if (!hashSet.has(num)) {
            return false;
        }
    }

    // If all elements of b are found in the hash set
    return true;
}

// Driver code
const a = [11, 1, 13, 21, 3, 7];
const b = [11, 3, 7, 1];

if (isSubset(a, b)) {
    console.log("true");
} else {
    console.log("false");
}

Output

true

Time Complexity: O(m + n), where m and n are the size of a and b respectively.
Auxiliary Space: O(m)

Union and Intersection of two Linked List using Hashing

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