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[]

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

 // C++ program to find whether an array // is subset of another array #include    /* 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; }

 // 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[]");      } }

 # 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

 // 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



Output:
arr2[] is subset of arr1[]

Time Complexity: O(m*n)

Method 2 (Use Sorting and Binary Search):

• Sort arr1[] which takes O(mLogm)
• For each element of arr2[], do binary search for it in sorted arr1[].
• If all elements are present then return 1.
 // C++ program to find whether an array // is subset of another array #include 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= 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

 // C program to find whether an array // is subset of another array #include #include /* 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= 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; }

 // 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= 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 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[]");       } }

 # 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

 // 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= 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 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

 = \$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 ?>

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 )

• Sort both arrays: arr1[] and arr2[] which takes O(mLogm + nLogn)
• Use Merge type of process to see if all elements of sorted arr2[] are present in sorted arr1[].

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:

 // C++ program to find whether an array // is subset of another array #include 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] )             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; }

 // 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

 # 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

 // 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.

 \$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. ?>

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)

• Create a Hash Table for all the elements of arr1[].
• Traverse arr2[] and search for each element of arr2[] in the Hash Table. If element is not found then return 0.
• If all elements are found then return 1.

Below is the implementation of the above approach:

 // 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 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

 // 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 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; }    // 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

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.