Find whether an array is subset of another array | Added Method 3

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 array 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[]

Method 1 (Simple):
Use two loops: The outer loop picks all the elements of arr2[] one by one. The inner loop linearly searches for the element picked by the outer loop. If all elements are found then return 1, else return 0.

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// C++ program to find whether an array
// is subset of another array
#include<bits/stdc++.h>
  
/* Return 1 if arr2[] is a subset of 
arr1[] */
bool isSubset(int arr1[], int arr2[], 
                        int m, int n)
{
    int i = 0;
    int j = 0;
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < m; j++)
        {
            if(arr2[i] == arr1[j])
                break;
        }
          
        /* If the above inner loop was
        not broken at all then arr2[i]
        is not present in arr1[] */
        if (j == m)
            return 0;
    }
      
    /* If we reach here then all
    elements of arr2[] are present
    in arr1[] */
    return 1;
}
  
// 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))
        printf("arr2[] is subset of arr1[] ");
    else
        printf("arr2[] is not a subset of arr1[]");     
  
    getchar();
    return 0;
}
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// Java program to find whether an array
// is subset of another array
  
class GFG {
  
    /* Return true if arr2[] is a subset 
    of arr1[] */
    static boolean isSubset(int arr1[], 
                int arr2[], int m, int n)
    {
        int i = 0;
        int j = 0;
        for (i = 0; i < n; i++)
        {
            for (j = 0; j < m; j++)
                if(arr2[i] == arr1[j])
                    break;
              
            /* If the above inner loop 
            was not broken at all then
            arr2[i] is not present in
            arr1[] */
            if (j == m)
                return false;
        }
          
        /* If we reach here then all
        elements of arr2[] are present
        in arr1[] */
        return true;
    }
      
    // 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))
            System.out.print("arr2[] is "
                  + "subset of arr1[] ");
        else
            System.out.print("arr2[] is "
             + "not a subset of arr1[]"); 
    }
}
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# Python 3 program to find whether an array
# is subset of another array
  
# Return 1 if arr2[] is a subset of 
# arr1[] 
def isSubset(arr1, arr2, m, n):
    i = 0
    j = 0
    for i in range(n):
        for j in range(m):
            if(arr2[i] == arr1[j]):
                break
          
        # If the above inner loop was
        # not broken at all then arr2[i]
        # is not present in arr1[] 
        if (j == m):
            return 0
      
    # If we reach here then all
    # elements of arr2[] are present
    # in arr1[] 
    return 1
  
# Driver code
if __name__ == "__main__":
      
    arr1 = [11, 1, 13, 21, 3, 7]
    arr2 = [11, 3, 7, 1]
  
    m = len(arr1)
    n = len(arr2)
  
    if(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 ita_c
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// C# program to find whether an array
// is subset of another array
using System;
  
class GFG {
  
    /* Return true if arr2[] is a 
    subset of arr1[] */
    static bool isSubset(int []arr1, 
               int []arr2, int m, int n)
    {
        int i = 0;
        int j = 0;
        for (i = 0; i < n; i++)
        {
            for (j = 0; j < m; j++)
                if(arr2[i] == arr1[j])
                    break;
              
            /* If the above inner loop 
            was not broken at all then
            arr2[i] is not present in
            arr1[] */
            if (j == m)
                return false;
        }
          
        /* If we reach here then all
        elements of arr2[] are present
        in arr1[] */
        return true;
    }
      
    // Driver function
    public static void Main()
    {
        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))
        Console.WriteLine("arr2[] is subset"
                           + " of arr1[] ");
        else
        Console.WriteLine("arr2[] is not a "
                      + "subset of arr1[]");
    }
}
  
// This code is contributed by Sam007
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<?php
// PHP program to find whether an array
// is subset of another array
  
/* Return 1 if arr2[] is a subset of 
arr1[] */
function isSubset($arr1, $arr2, $m, $n)
{
    $i = 0;
    $j = 0;
    for ($i = 0; $i < $n; $i++)
    {
        for ($j = 0; $j < $m; $j++)
        {
            if($arr2[$i] == $arr1[$j])
                break;
        }
          
        /* If the above inner loop was
        not broken at all then arr2[i]
        is not present in arr1[] */
        if ($j == $m)
            return 0;
    }
      
    /* If we reach here then all
    elements of arr2[] are present
    in arr1[] */
    return 1;
}
  
// Driver code
    $arr1 = array(11, 1, 13, 21, 3, 7);
    $arr2 = array(11, 3, 7, 1);
  
    $m = count($arr1);
    $n = count($arr2);
  
    if(isSubset($arr1, $arr2, $m, $n))
        echo "arr2[] is subset of arr1[] ";
    else
        echo "arr2[] is not a subset of arr1[]";     
  
// This code is contributed by anuj_67.
?>
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Output:
arr2[] is subset of arr1[] 

Time Complexity: O(m*n)

Method 2 (Use Sorting and Binary Search):

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// C++ program to find whether an array
// is subset of another array
#include<bits/stdc++.h>
using namespace std;
  
/* Fucntion prototypes */
void quickSort(int *arr, int si, int ei);
int binarySearch(int arr[], int low, 
                    int high, int x);
  
/* Return 1 if arr2[] is a subset of arr1[] */
bool isSubset(int arr1[], int arr2[],
                        int m, int n)
{
    int i = 0;
  
    quickSort(arr1, 0, m-1);
    for (i=0; i<n; i++)
    {
        if (binarySearch(arr1, 0, m - 1,
                        arr2[i]) == -1)
        return 0;
    }
      
    /* If we reach here then all elements
     of arr2[] are present in arr1[] */
    return 1;
}
  
/* FOLLOWING FUNCTIONS ARE ONLY FOR 
    SEARCHING AND SORTING PURPOSE */
/* Standard Binary Search function*/
int binarySearch(int arr[], int low,
                    int high, int x)
{
    if(high >= low)
    {
        int mid = (low + high)/2; /*low + (high - low)/2;*/
  
        /* Check if arr[mid] is the first occurrence of x.
        arr[mid] is first occurrence if x is one of the following
        is true:
        (i) mid == 0 and arr[mid] == x
        (ii) arr[mid-1] < x and arr[mid] == x    */
        if(( mid == 0 || x > arr[mid-1]) && (arr[mid] == x))
            return mid;
        else if(x > arr[mid])
            return binarySearch(arr, (mid + 1), high, x);
        else
            return binarySearch(arr, low, (mid -1), x);
    }
    return -1;
  
void exchange(int *a, int *b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}
  
int partition(int A[], int si, int ei)
{
    int x = A[ei];
    int i = (si - 1);
    int j;
  
    for (j = si; j <= ei - 1; j++)
    {
        if(A[j] <= x)
        {
            i++;
            exchange(&A[i], &A[j]);
        }
    }
    exchange (&A[i + 1], &A[ei]);
    return (i + 1);
}
  
/* Implementation of Quick Sort
A[] --> Array to be sorted
si --> Starting index
ei --> Ending index
*/
void quickSort(int A[], int si, int ei)
{
    int pi; /* Partitioning index */
    if(si < ei)
    {
        pi = partition(A, si, ei);
        quickSort(A, si, pi - 1);
        quickSort(A, pi + 1, ei);
    }
}
  
/*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;
}
  
// This code is contributed by Shivi_Aggarwal
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// C program to find whether an array
// is subset of another array
#include<stdio.h>
#include <stdbool.h>
/* Fucntion prototypes */
void quickSort(int *arr, int si, int ei);
int binarySearch(int arr[], int low, int high, int x);
  
/* Return 1 if arr2[] is a subset of arr1[] */
bool isSubset(int arr1[], int arr2[], int m, int n)
{
    int i = 0;
    
    quickSort(arr1, 0, m-1);
    for (i=0; i<n; i++)
    {
        if (binarySearch(arr1, 0, m-1, arr2[i]) == -1)
           return 0;
    }
      
    /* If we reach here then all elements of arr2[] 
      are present in arr1[] */
    return 1;
}
   
/* FOLLOWING FUNCTIONS ARE ONLY FOR SEARCHING AND SORTING PURPOSE */
/* Standard Binary Search function*/
int binarySearch(int arr[], int low, int high, int x)
{
  if(high >= low)
  {
    int mid = (low + high)/2;  /*low + (high - low)/2;*/
   
    /* Check if arr[mid] is the first occurrence of x.
        arr[mid] is first occurrence if x is one of the following
        is true:
        (i)  mid == 0 and arr[mid] == x
        (ii) arr[mid-1] < x and arr[mid] == x
     */
    if(( mid == 0 || x > arr[mid-1]) && (arr[mid] == x))
      return mid;
    else if(x > arr[mid])
      return binarySearch(arr, (mid + 1), high, x);
    else
      return binarySearch(arr, low, (mid -1), x);
  }
 return -1;
}  
  
void exchange(int *a, int *b)
{
    int temp;
    temp = *a;
    *a   = *b;
    *b   = temp;
}
   
int partition(int A[], int si, int ei)
{
    int x = A[ei];
    int i = (si - 1);
    int j;
   
    for (j = si; j <= ei - 1; j++)
    {
        if(A[j] <= x)
        {
            i++;
            exchange(&A[i], &A[j]);
        }
    }
    exchange (&A[i + 1], &A[ei]);
    return (i + 1);
}
   
/* Implementation of Quick Sort
A[] --> Array to be sorted
si  --> Starting index
ei  --> Ending index
*/
void quickSort(int A[], int si, int ei)
{
    int pi;    /* Partitioning index */
    if(si < ei)
    {
        pi = partition(A, si, ei);
        quickSort(A, si, pi - 1);
        quickSort(A, pi + 1, ei);
    }
}
   
/*Driver program to test above functions */
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))
      printf("arr2[] is subset of arr1[] ");
    else
      printf("arr2[] is not a subset of arr1[] ");      
  
    return 0;
}
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// Java program to find whether an array
// is subset of another array
class Main
{
    /* Return true if arr2[] is a subset of arr1[] */
    static boolean isSubset(int arr1[], int arr2[], int m, int n)
    {
        int i = 0;
         
        sort(arr1, 0, m-1);
        for (i=0; i<n; i++)
        {
            if (binarySearch(arr1, 0, m-1, arr2[i]) == -1)
               return false;
        }
           
        /* If we reach here then all elements of arr2[] 
          are present in arr1[] */
        return true;
    }
        
    /* FOLLOWING FUNCTIONS ARE ONLY FOR SEARCHING AND SORTING PURPOSE */
    /* Standard Binary Search function*/
    static int binarySearch(int arr[], int low, int high, int x)
    {
      if(high >= low)
      {
        int mid = (low + high)/2/*low + (high - low)/2;*/
        
        /* Check if arr[mid] is the first occurrence of x.
            arr[mid] is first occurrence if x is one of the following
            is true:
            (i)  mid == 0 and arr[mid] == x
            (ii) arr[mid-1] < x and arr[mid] == x
         */
        if(( mid == 0 || x > arr[mid-1]) && (arr[mid] == x))
          return mid;
        else if(x > arr[mid])
          return binarySearch(arr, (mid + 1), high, x);
        else
          return binarySearch(arr, low, (mid -1), x);
      }
     return -1;
    }  
       
    /* This function takes last element as pivot,
       places the pivot element at its correct
       position in sorted array, and places all
       smaller (smaller than pivot) to left of
       pivot and all greater elements to right
       of pivot */
    static int partition(int arr[], int low, int high)
    {
        int pivot = arr[high]; 
        int i = (low-1); // index of smaller element
        for (int j=low; j<high; j++)
        {
            // If current element is smaller than or
            // equal to pivot
            if (arr[j] <= pivot)
            {
                i++;
   
                // swap arr[i] and arr[j]
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
   
        // swap arr[i+1] and arr[high] (or pivot)
        int temp = arr[i+1];
        arr[i+1] = arr[high];
        arr[high] = temp;
   
        return i+1;
    }
   
   
    /* The main function that implements QuickSort()
      arr[] --> Array to be sorted,
      low  --> Starting index,
      high  --> Ending index */
    static void sort(int arr[], int low, int high)
    {
        if (low < high)
        {
            /* pi is partitioning index, arr[pi] is 
              now at right place */
            int pi = partition(arr, low, high);
   
            // Recursively sort elements before
            // partition and after partition
            sort(arr, low, pi-1);
            sort(arr, pi+1, high);
        }
    }
       
    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))
          System.out.print("arr2[] is subset of arr1[] ");
        else
          System.out.print("arr2[] is not a subset of arr1[]");  
    }
}
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# Python3 program to find whether an array
# is subset of another array
  
# Return 1 if arr2[] is a subset of arr1[] 
def isSubset(arr1, arr2, m, n):
    i = 0;
  
    quickSort(arr1, 0, m-1);
    for i in range(n):
        if (binarySearch(arr1, 0, m - 1,arr2[i]) == -1):
            return 0;
      
    # If we reach here then all elements
    # of arr2[] are present in arr1[] 
    return 1;
  
# FOLLOWING FUNCTIONS ARE ONLY FOR 
# SEARCHING AND SORTING PURPOSE
# Standard Binary Search function
def binarySearch(arr, low, high, x):
    if(high >= low):
        mid = (low + high)//2;
  
        # Check if arr[mid] is the first occurrence of x.
        # arr[mid] is first occurrence if x is one of the following
        # is true:
        # (i) mid == 0 and arr[mid] == x
        # (ii) arr[mid-1] < x and arr[mid] == x 
        if(( mid == 0 or x > arr[mid-1]) and (arr[mid] == x)):
            return mid;
        elif(x > arr[mid]):
            return binarySearch(arr, (mid + 1), high, x);
        else:
            return binarySearch(arr, low, (mid -1), x);
  
    return -1;
  
  
def partition(A, si, ei):
    x = A[ei];
    i = (si - 1);
  
    for j in range(si,ei):
        if(A[j] <= x):
            i+=1;
            A[i],A[j] = A[j],A[i];
    A[i + 1],A[ei] = A[ei],A[i+1];
    return (i + 1);
  
# Implementation of Quick Sort
# A[] --> Array to be sorted
# si --> Starting index
# ei --> Ending index
  
def quickSort(A, si, ei):
    # Partitioning index 
    if(si < ei):
        pi = partition(A, si, ei);
        quickSort(A, si, pi - 1);
        quickSort(A, pi + 1, ei);
  
# Driver code 
arr1 = [11, 1, 13, 21, 3, 7];
arr2 = [11, 3, 7, 1];
  
m = len(arr1);
n = len(arr2);
  
if(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 chandan_jnu
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// C# program to find whether an array
// is subset of another array
using System;
                  
      
public class GFG
{
    /* Return true if arr2[] is a subset of arr1[] */
    static bool isSubset(int []arr1, int []arr2, int m, int n)
    {
        int i = 0;
          
        sort(arr1, 0, m-1);
        for (i=0; i<n; i++)
        {
            if (binarySearch(arr1, 0, m-1, arr2[i]) == -1)
               return false;
        }
            
        /* If we reach here then all elements of arr2[] 
          are present in arr1[] */
        return true;
    }
         
    /* FOLLOWING FUNCTIONS ARE ONLY FOR SEARCHING AND SORTING PURPOSE */
    /* Standard Binary Search function*/
    static int binarySearch(int []arr, int low, int high, int x)
    {
      if(high >= low)
      {
        int mid = (low + high)/2;  /*low + (high - low)/2;*/
         
        /* Check if arr[mid] is the first occurrence of x.
            arr[mid] is first occurrence if x is one of the following
            is true:
            (i)  mid == 0 and arr[mid] == x
            (ii) arr[mid-1] < x and arr[mid] == x
         */
        if(( mid == 0 || x > arr[mid-1]) && (arr[mid] == x))
          return mid;
        else if(x > arr[mid])
          return binarySearch(arr, (mid + 1), high, x);
        else
          return binarySearch(arr, low, (mid -1), x);
      }
     return -1;
    }  
        
    /* This function takes last element as pivot,
       places the pivot element at its correct
       position in sorted array, and places all
       smaller (smaller than pivot) to left of
       pivot and all greater elements to right
       of pivot */
    static int partition(int []arr, int low, int high)
    {
        int pivot = arr[high]; 
        int i = (low-1); // index of smaller element
        int temp=0;
        for (int j=low; j<high; j++)
        {
            // If current element is smaller than or
            // equal to pivot
            if (arr[j] <= pivot)
            {
                i++;
    
                // swap arr[i] and arr[j]
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    
        // swap arr[i+1] and arr[high] (or pivot)
        temp = arr[i+1];
        arr[i+1] = arr[high];
        arr[high] = temp;
    
        return i+1;
    }
    
    
    /* The main function that implements QuickSort()
      arr[] --> Array to be sorted,
      low  --> Starting index,
      high  --> Ending index */
    static void sort(int []arr, int low, int high)
    {
        if (low < high)
        {
            /* pi is partitioning index, arr[pi] is 
              now at right place */
            int pi = partition(arr, low, high);
    
            // Recursively sort elements before
            // partition and after partition
            sort(arr, low, pi-1);
            sort(arr, pi+1, high);
        }
    }
        
    public static void Main()
    {
        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))
          Console.Write("arr2[] is subset of arr1[] ");
        else
          Console.Write("arr2[] is not a subset of arr1[]");  
    }
}
//This code is contributed by 29AjayKumar
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<?php
// PHP program to find whether an array
// is subset of another array
  
/* Return 1 if arr2[] is a subset of arr1[] */
function isSubset($arr1, $arr2,
                        $m, $n)
{
    $i = 0;
  
    quickSort($arr1, 0, $m-1);
    for ($i = 0; $i < $n; $i++)
    {
        if (binarySearch($arr1, 0, $m - 1,
                        $arr2[$i]) == -1)
        return 0;
    }
      
    /* If we reach here then all elements
    of arr2[] are present in arr1[] */
    return 1;
}
  
/* FOLLOWING FUNCTIONS ARE ONLY FOR 
    SEARCHING AND SORTING PURPOSE */
/* Standard Binary Search function*/
function binarySearch($arr, $low, $high, $x)
{
    if($high >= $low)
    {
        $mid = (int)(($low + $high)/2); /*low + (high - low)/2;*/
  
        /* Check if arr[mid] is the first occurrence of x.
        arr[mid] is first occurrence if x is one of the following
        is true:
        (i) mid == 0 and arr[mid] == x
        (ii) arr[mid-1] < x and arr[mid] == x */
        if(( $mid == 0 || $x > $arr[$mid-1]) && ($arr[$mid] == $x))
            return $mid;
        else if($x > $arr[$mid])
            return binarySearch($arr, ($mid + 1), $high, $x);
        else
            return binarySearch($arr, $low, ($mid -1), $x);
    }
    return -1;
  
function exchange(&$a, &$b)
{
  
    $temp = $a;
    $a = $b;
    $b = $temp;
}
  
function partition(&$A, $si, $ei)
{
    $x = $A[$ei];
    $i = ($si - 1);
  
    for ($j = $si; $j <= $ei - 1; $j++)
    {
        if($A[$j] <= $x)
        {
            $i++;
            exchange($A[$i], $A[$j]);
        }
    }
    exchange ($A[$i + 1], $A[$ei]);
    return ($i + 1);
}
  
/* Implementation of Quick Sort
A[] --> Array to be sorted
si --> Starting index
ei --> Ending index
*/
function quickSort(&$A, $si, $ei)
{
    /* Partitioning index */
    if($si < $ei)
    {
        $pi = partition($A, $si, $ei);
        quickSort($A, $si, $pi - 1);
        quickSort($A, $pi + 1, $ei);
    }
}
  
    /*Driver code */
    $arr1 = array(11, 1, 13, 21, 3, 7);
    $arr2 = array(11, 3, 7, 1);
  
    $m = count($arr1);
    $n = count($arr2);
  
    if(isSubset($arr1, $arr2, $m, $n))
        echo "arr2[] is subset of arr1[] ";
    else
        echo "arr2[] is not a subset of arr1[] "
  
  
// This code is contributed by chandan_jnu
?>
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Output:
arr2 is a subset of arr1

Time Complexity: O(mLogm + nLogm). Please note that this will be the complexity if an mLogm algorithm is used for sorting which is not the case in above code. In above code Quick Sort is used and worst case time complexity of Quick Sort is O(m^2)

Method 3 (Use Sorting and Merging )

Thanks to Parthsarthi for suggesting this method.


Below image is a dry run of the above approach:

Below is the implementation of the above approach:

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// C++ program to find whether an array
// is subset of another array
#include <bits/stdc++.h>
using namespace std;
  
/* Return 1 if arr2[] is a subset of arr1[] */ 
bool isSubset(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
      
    if (m < n)
       return 0;
      
    // Sort both the arrays 
    sort(arr1, arr1 + m);
    sort(arr2, arr2 + n); 
      
    // Iterate till they donot exceed their sizes 
    while (i < n && j < m )
    
        // If the element is smaller than
        // Move aheaad in the first array 
        if( arr1[j] <arr2[i] )
            j++;
        // If both are equal, then move both of them forward 
        else if( arr1[j] == arr2[i] )
        {
            j++;
            i++;
        }
  
        // If we donot have a element smaller 
        // or equal to the second array then break 
        else if( arr1[j] > arr2[i] )
            return 0;
    }
   
    return  (i < n)? false : true;
  
// 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))
      printf("arr2[] is subset of arr1[] ");
    else
      printf("arr2[] is not a subset of arr1[] ");      
  
    return 0;
}
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// Java code to find whether an array is subset of 
// another array
import java.util.Arrays;
class GFG
{
    /* Return true if arr2[] is a subset of arr1[] */
    static boolean isSubset(int arr1[], int arr2[], int m,
                                                   int n)
    {
        int i = 0, j = 0;
              
        if(m < n)
        return false;
          
        Arrays.sort(arr1); //sorts arr1
        Arrays.sort(arr2); // sorts arr2
  
        while( i < n && j < m )
        {
            if( arr1[j] < arr2[i] )
                j++;
            else if( arr1[j] == arr2[i] )
            {
                j++;
                i++;
            }
            else if( arr1[j] > arr2[i] )
                return false;
        }
          
        if( i < n )
            return false;
        else
            return true;
    
          
    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))
        System.out.println("arr2 is a subset of arr1");
        else
        System.out.println("arr2 is not a subset of arr1");
    }
}
// This code is contributed by Kamal Rawal
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# Python3 program to find whether an array 
# is subset of another array 
  
# Return 1 if arr2[] is a subset of arr1[] */ 
def isSubset(arr1, arr2, m, n):
    i = 0
    j = 0
    if m < n:
        return 0
          
    arr1.sort()
    arr2.sort()
  
    while i < n and j < m:
        if arr1[j] < arr2[i]:
            j += 1
        elif arr1[j] == arr2[i]:
            j += 1
            i += 1
        elif arr1[j] > arr2[i]:
            return 0
    return False if i < n else True
  
# Driver code
arr1 = [11, 1, 13, 21, 3, 7]
arr2 = [11, 3, 7, 1]
  
m = len(arr1)
n = len(arr2)
if isSubset(arr1, arr2, m, n) == True:
    print("arr2 is subset of arr1 ")
else:
    printf("arr2 is not a subset of arr1 ")
  
# This code is contributed by Shrikant13
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// C# code to find whether an array
// is subset of another array
using System;
class GFG {
      
    // Return true if arr2[] is 
    // a subset of arr1[] */
    static bool isSubset(int []arr1, 
                         int []arr2, 
                         int m,
                         int n)
    {
        int i = 0, j = 0;
              
        if(m < n)
            return false;
          
        //sorts arr1
        Array.Sort(arr1); 
          
        // sorts arr2
        Array.Sort(arr2); 
  
        while( i < n && j < m )
        {
            if( arr1[j] < arr2[i] )
                j++;
            else if( arr1[j] == arr2[i] )
            {
                j++;
                i++;
            }
            else if( arr1[j] > arr2[i] )
                return false;
        }
          
        if( i < n )
            return false;
        else
            return true;
    
          
    // Driver Code
    public static void Main() 
    
        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))
            Console.Write("arr2 is a subset of arr1");
        else
            Console.Write("arr2 is not a subset of arr1");
    }
}
  
// This code is contributed by nitin mittal.
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<?php
// PHP program to find whether an array
// is subset of another array
  
/* Return 1 if arr2[] is a subset of arr1[] */
function isSubset( $arr1, $arr2, $m, $n)
{
    $i = 0; $j = 0;
      
    if ($m < $n)
        return 0;
  
    sort($arr1);
    sort($arr2);
      
    while ($i < $n and $j < $m )
    {
        if( $arr1[$j] <$arr2[$i] )
            $j++;
        else if( $arr1[$j] == $arr2[$i] )
        {
            $j++;
            $i++;
        }
        else if( $arr1[$j] > $arr2[$i] )
            return 0;
    }
  
    return ($i < $n) ? false : true;
  
/*Driver program to test above functions */
  
    $arr1 = array(11, 1, 13, 21, 3, 7);
    $arr2 = array(11, 3, 7, 1);
  
    $m = count($arr1);
    $n = count($arr2);
  
    if(isSubset($arr1, $arr2, $m, $n))
        echo "arr2[] is subset of arr1[] ";
    else
        echo "arr2[] is not a subset of arr1[] "
  
// This code is contributed by anuj_67.
?>
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Output:
arr2 is a subset of arr1

Time Complexity: O(mLogm + nLogn) which is better than method 2. Please note that this will be the complexity if an nLogn algorithm is used for sorting both arrays which is not the case in above code. In above code Quick Sort is used and worst case time complexity of Quick Sort is O(n^2)

Method 4 (Use Hashing)

Below is the implementation of the above approach:

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// Java code to find whether an array is subset of
// another array
import java.util.HashSet;
class GFG
{
    /* Return true if arr2[] is a subset of arr1[] */
    static boolean isSubset(int arr1[], int arr2[], int m,
                                                 int n)
    {
        HashSet<Integer> hset= new HashSet<>();
          
        // hset stores all the values of arr1
        for(int i = 0; i < m; i++)
        {
            if(!hset.contains(arr1[i]))
                hset.add(arr1[i]);
        }
              
        // loop to check if all elements of arr2 also
        // lies in arr1
        for(int i = 0; i < n; i++)
        {
            if(!hset.contains(arr2[i]))
                return false;
        }
        return true;
    
  
    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))
        System.out.println("arr2 is a subset of arr1");
        else
        System.out.println("arr2 is not a subset of arr1");
    }
}
// This code is contributed by Kamal Rawal
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// C# code to find whether an array is 
// subset of another array 
using System;
using System.Collections.Generic;
  
class GFG
{
/* Return true if arr2[] is a 
   subset of arr1[] */
public static bool isSubset(int[] arr1, 
                            int[] arr2,
                            int m, int n)
{
    HashSet<int> hset = new HashSet<int>();
  
    // hset stores all the values of arr1 
    for (int i = 0; i < m; i++)
    {
        if (!hset.Contains(arr1[i]))
        {
            hset.Add(arr1[i]);
        }
    }
  
    // loop to check if all elements 
    // of arr2 also lies in arr1 
    for (int i = 0; i < n; i++)
    {
        if (!hset.Contains(arr2[i]))
        {
            return false;
        }
    }
    return true;
}
  
// Driver Code
public static void Main(string[] args)
{
    int[] arr1 = new int[] {11, 1, 13, 21, 3, 7};
    int[] arr2 = new int[] {11, 3, 7, 1};
  
    int m = arr1.Length;
    int n = arr2.Length;
  
    if (isSubset(arr1, arr2, m, n))
    {
        Console.WriteLine("arr2 is a subset of arr1");
    }
    else
    {
        Console.WriteLine("arr2 is not a subset of arr1");
    }
}
}
  
// This code is contributed by Shrikant13
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Output:
arr2 is a subset of arr1

Note that method 1, method 2 and method 4 don’t handle the cases when we have duplicates in arr2[]. For example, {1, 4, 4, 2} is not a subset of {1, 4, 2}, but these methods will print it as a subset.

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.






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