Open in App
Not now

# Find whether an array is subset of another array

• Difficulty Level : Easy
• Last Updated : 01 Mar, 2023

Given two arrays: arr1[0..m-1] and arr2[0..n-1]. Find whether arr2[] is a subset of arr1[] or not. Both 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[]

Recommended Practice

## Naive Approach to Find whether an array is subset of another array

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.

Below is the implementation of the above approach:

## C++

 // 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; }

## C

 // C++ program to find whether an array // is subset of another array #include /* Return 1 if arr2[] is a subset of arr1[] */ int 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

 // 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[]");     } }

## Python3

 # 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#

 // 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



## Javascript



Output

arr2[] is subset of arr1[]

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

## Find whether an array is subset of another array using Sorting and Binary Search

The idea is to sort the given array arr1[], and then for each element in arr2[] do a binary search for it in sorted arr1[]. If the element is not found then return 0. If all elements are present then return 1.

Illustration:

Given array arr1[] = { 11, 1, 13, 21, 3, 7 } and arr2[] = { 11, 3, 7, 1 }.

Step 1: We will sort the array arr1[], and have arr1[] = { 1, 3, 7, 11, 13, 21}.

Step 2: We will look for each element in arr2[] in arr1[] using binary search.

• arr2[] = { 11, 3, 7, 1 }, 11 is present in arr1[] = { 1, 3, 7, 11, 13, 21}
• arr2[] = { 11, 3, 7, 1 }, 3 is present in arr1[] = { 1, 3, 7, 11, 13, 21}
• arr2[] = { 11, 3, 7, 1 }, 7 is present in arr1[] = { 1, 3, 7, 11, 13, 21}
• arr2[] = { 11, 3, 7, 1 }, 1 is present in arr1[] = { 1, 3, 7, 11, 13, 21}

As all the elements are found we can conclude arr2[] is the subset of arr1[].

Algorithm:

The algorithm is pretty straightforward.

• Sort the first array arr1[].
• Look for the elements of arr2[] in sorted arr1[].
• If we encounter a particular value that is present in arr2[] but not in arr1[], the code will terminate, arr2[] can never be the subset of arr1[].
• Else arr2[] is the subset of arr1[].

Below is the code implementation of the above approach :

## C++

 // C++ program to find whether an array // is subset of another array #include using namespace std;   /* Function 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) {         /*low + (high - low)/2;*/         int 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 || 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

 // C program to find whether an array // is subset of another array #include #include /* Function 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) {         /*low + (high - low)/2;*/         int 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 || 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))         printf("arr2[] is subset of arr1[] ");     else         printf("arr2[] is not a subset of arr1[] ");       return 0; }

## Java

 // 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) {             /*low + (high - low)/2;*/             int 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 || 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);           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);         }     }       // 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[]");     } }

## Python3

 # 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#

 // 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;               /* 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);         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);         }     }       // 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 subset of arr1[] ");         else             Console.Write(                 "arr2[] is not a subset of arr1[]");     } } // This code is contributed by 29AjayKumar

## PHP

 = \$low)     {         \$mid = (int)((\$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 || \$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 ?>

## Javascript



Output

arr2[] is subset of arr1[]

Time Complexity: O(mLog(m) + nlog(m)). O(mLog(m)) for sorting and O(nlog(m)) for binary searching each element of one array in another. In the above code, Quick Sort is used and the worst-case time complexity of Quick Sort is O(m2).
Auxiliary Space: O(n)

## Find whether an array is subset of another array using Sorting and Merging

The idea is to sort the two arrays and then iterate on the second array looking for the same values on the first array using two pointers. Whenever we encounter the same values we will increment both the pointer and if we encounter any values less than that of the second array, we will increment the value of the pointer pointing to the first array. If the value is greater than that of the second array, we know the second array is not the subset of the first array.

Illustration:

Algorithm:

The initial step will be to sort the two arrays.

• Set two pointers j and i for arr1[] and arr2[] respectively.
• If arr1[j] < arr2[i], we will increase j by 1.
• If arr1[j] = arr2[i], we will increase j and i by 1.
• If arr1[j] > arr2[i], we will terminate as arr2[] is not the subset of arr1[].

Below is the implementation of the above approach:

## C++

 // 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 do not exceed their sizes     while (i < n && j < m) {         // If the element is smaller then         // Move ahead 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 do not have an 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; }

## C

 // C program to find whether an array // is subset of another array #include #include   // Merges two subarrays of arr[]. // First subarray is arr[l..m] // Second subarray is arr[m+1..r] void merge(int arr[], int l, int m, int r) {     int i, j, k;     int n1 = m - l + 1;     int n2 = r - m;       /* create temp arrays */     int L[n1], R[n2];       /* Copy data to temp arrays L[] and R[] */     for (i = 0; i < n1; i++)         L[i] = arr[l + i];     for (j = 0; j < n2; j++)         R[j] = arr[m + 1 + j];       /* Merge the temp arrays back into arr[l..r]*/     i = 0; // Initial index of first subarray     j = 0; // Initial index of second subarray     k = l; // Initial index of merged subarray     while (i < n1 && j < n2) {         if (L[i] <= R[j]) {             arr[k] = L[i];             i++;         }         else {             arr[k] = R[j];             j++;         }         k++;     }       /* Copy the remaining elements of L[], if there     are any */     while (i < n1) {         arr[k] = L[i];         i++;         k++;     }       /* Copy the remaining elements of R[], if there     are any */     while (j < n2) {         arr[k] = R[j];         j++;         k++;     } }   /* l is for left index and r is right index of the sub-array of arr to be sorted */ void mergeSort(int arr[], int l, int r) {     if (l < r) {         // Same as (l+r)/2, but avoids overflow for         // large l and h         int m = l + (r - l) / 2;           // Sort first and second halves         mergeSort(arr, l, m);         mergeSort(arr, m + 1, r);           merge(arr, l, m, r);     } } /* Return 1 if arr2[] is a subset of arr1[] */ int isSubset(int arr1[], int arr2[], int m, int n) {     int i = 0, j = 0;       if (m < n)         return 0;       // Sort both the arrays     mergeSort(arr1, 0, m - 1);     mergeSort(arr2, 0, n - 1);       // Iterate till they do not exceed their sizes     while (i < n && j < m) {         // If the element is smaller then         // Move ahead 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 do not have an element smaller         // or equal to the second array then break         else if (arr1[j] > arr2[i])             return 0;     }       return (i < n) ? 0 : 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[] ");       return 0; }

## Java

 // 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;     }       // 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.println("arr2[] is subset of arr1[] ");         else             System.out.println(                 "arr2[] is not a subset of arr1[] ");     } } // This code is contributed by Kamal Rawal

## Python3

 # 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#

 // 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 subset of arr1[] ");         else             Console.Write("arr2[] is not a subset of arr1[] ");     } }   // This code is contributed by nitin mittal.

## PHP

 \$arr2[\$i] )             return 0;     }       return (\$i < \$n) ? false : true; }   /*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. ?>

## Javascript



Output

arr2[] is subset of arr1[]

Time Complexity: O(mLog(m) + nLog(n)) which is better than approach 2.
Auxiliary Space: O(1)

Thanks to Parthsarthi for suggesting this method.

## Find whether an array is a subset of another array using Hashing

The idea is to insert all the elements of the first array in a HashSet, and then iterate on the second array and find if the element exists in the HashSet, if the HashSet doesn’t contain any particular value then the second array is not the subset of the first array.

Note: This approach works perfectly if there are no duplicate elements.

Illustration:

Given array arr1[] = { 11, 1, 13, 21, 3, 7 } and arr2[] = { 11, 3, 7, 1 }.

Step 1: We will store the array arr1[] elements in HashSet

Step 2: We will look for each element in arr2[] in arr1[] using binary search.

• arr2[] = { 11, 3, 7, 1 }, 11 is present in the HashSet = { 1, 3, 7, 11, 13, 21}
• arr2[] = { 11, 3, 7, 1 }, 3 is present in the HashSet = { 1, 3, 7, 11, 13, 21}
• arr2[] = { 11, 3, 7, 1 }, 7 is present in the HashSet = { 1, 3, 7, 11, 13, 21}
• arr2[] = { 11, 3, 7, 1 }, 1 is present in the HashSet = { 1, 3, 7, 11, 13, 21}

As all the elements are found we can conclude arr2[] is the subset of arr1[].

Algorithm:

The algorithm is pretty straightforward.

• Store the first array arr1[] in a HashSet.
• Look for the elements of arr2[] in the HashSet.
• If we encounter a particular value that is present in arr2[] but not in the HashSet, the code will terminate, arr2[] can never be the subset of arr1[].
• Else arr2[] is the subset of arr1[].

Below is the implementation of the above approach:

## C++

 // C++ code to find whether an array is subset of // another array #include using namespace std;   /* Return true if arr2[] is a subset of arr1[] */ bool isSubset(int arr1[], int m, int arr2[], int n) {       // Using STL set for hashing     set hashset;       // hashset stores all the values of arr1     for (int i = 0; i < m; i++) {         hashset.insert(arr1[i]);     }       // loop to check if all elements of arr2 also     // lies in arr1     for (int i = 0; i < n; i++) {         if (hashset.find(arr2[i]) == hashset.end())             return false;     }     return 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, m, arr2, n))         cout << "arr2[] is subset of arr1[] "              << "\n";     else         cout << "arr2[] is not a subset of arr1[] "              << "\n";     return 0; } // This code is contributed by Satvik Shrivas

## Java

 // 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;     }       // 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.println("arr2[] is subset of arr1[] ");         else             System.out.println(                 "arr2[] is not a subset of arr1[] ");     } } // This code is contributed by Kamal Rawal

## Python3

 # Python3 program to find whether an array # is subset of another array   # Return true if arr2[] is a subset # of arr1[]     def isSubset(arr1, m, arr2, n):       # Using STL set for hashing     hashset = set()       # hset stores all the values of arr1     for i in range(0, m):         hashset.add(arr1[i])       # Loop to check if all elements     # of arr2 also lies in arr1     for i in range(0, n):         if arr2[i] in hashset:             continue         else:             return False       return True     # 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, m, arr2, n)):         print("arr2[] is subset of arr1[] ")     else:         print("arr2[] is not a subset of arr1[] ")   # This code is contributed by akhilsaini

## C#

 # Python3 program to find whether an array # is subset of another array   # Return true if arr2[] is a subset # of arr1[]     def isSubset(arr1, m, arr2, n):       # Using STL set for hashing     hashset = set()       # hset stores all the values of arr1     for i in range(0, m):         hashset.add(arr1[i])       # Loop to check if all elements     # of arr2 also lies in arr1     for i in range(0, n):         if arr2[i] in hashset:             continue         else:             return False       return True     # 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, m, arr2, n)):         print("arr2[] is subset of arr1[] ")     else:         print("arr2[] is not a subset of arr1[] ")   # This code is contributed by akhilsaini



## PHP



Output

arr2[] is subset of arr1[]

Time Complexity: O(m+n*logm)
Auxiliary Space: O(m)

## Find whether an array is a subset of another array using Set

The idea is to insert all the elements of the first array and second array in the set, if the size of the set is equal to the size of arr1[] then the arr2[] is the subset of arr1[]. As no new elements are found in arr2[] hence is the subset.

Note: This approach works perfectly if there are no duplicate elements.

Illustration:

Given array arr1[] = { 11, 1, 13, 21, 3, 7 } and arr2[] = { 11, 3, 7, 1 }.

Step 1: We will store the array arr1[] and arr2[] elements in Set

• The final Set = { 1, 3, 7, 11, 13, 21}

Step 2: Size of arr1[] = 6 and size of the Set = 6

• Hence no new elements are found in arr2[]

As all the elements are found we can conclude arr2[] is the subset of arr1[].

Algorithm:

The algorithm is pretty straightforward.

• Store the first array arr1[] in a Set.
• Store the first array arr1[] in the same Set.
• If the size of arr1[] = size of the Set, arr2[] is the subset of arr1[].
• Else arr2[] is not the subset of arr1[].

## C++

 #include using namespace std;   int main() {     // code     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]);     unordered_set s;     for (int i = 0; i < m; i++) {         s.insert(arr1[i]);     }     int p = s.size();     for (int i = 0; i < n; i++) {         s.insert(arr2[i]);     }     if (s.size() == p) {         cout << "arr2[] is subset of arr1[] "              << "\n";     }     else {         cout << "arr2[] is not subset of arr1[] "              << "\n";     }     return 0; }

## Java

 import java.io.*; import java.util.*;   class GFG {     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;           Set s = new HashSet();         for (int i = 0; i < m; i++) {             s.add(arr1[i]);         }         int p = s.size();         for (int i = 0; i < n; i++) {             s.add(arr2[i]);         }           if (s.size() == p) {             System.out.println("arr2[] is subset of arr1[] "                                + "\n");         }         else {             System.out.println(                 "arr2[] is not subset of arr1[] "                 + "\n");         }     } }   // This code is contributed by avanitrachhadiya2155

## Python3

 # Python3 code arr1 = [11, 1, 13, 21, 3, 7] arr2 = [11, 3, 7, 1] m = len(arr1) n = len(arr2) s = set() for i in range(m):     s.add(arr1[i])   p = len(s) for i in range(n):     s.add(arr2[i])   if (len(s) == p):     print("arr2[] is subset of arr1[] ")   else:     print("arr2[] is not subset of arr1[] ")       # This code is contributed by divyeshrabadiya07.

## C#

 using System; using System.Collections.Generic;   public class GFG {     static public void Main()     {         int[] arr1 = { 11, 1, 13, 21, 3, 7 };         int[] arr2 = { 11, 3, 7, 1 };         int m = arr1.Length;         int n = arr2.Length;           HashSet s = new HashSet();         for (int i = 0; i < m; i++) {             s.Add(arr1[i]);         }         int p = s.Count;         for (int i = 0; i < n; i++) {             s.Add(arr2[i]);         }           if (s.Count == p) {             Console.WriteLine("arr2[] is subset of arr1[] "                               + "\n");         }         else {             Console.WriteLine(                 "arr2[] is not subset of arr1[] "                 + "\n");         }     } }   // This code is contributed by rag2127

## Javascript



Output

arr2[] is subset of arr1[]

Time Complexity: O(m+n) because we are using unordered_set and inserting in it, If we would be using an ordered set inserting would have taken log n increasing the TC to O(mlogm+nlogn), but order does not matter in this approach.
Auxiliary Space: O(n+m)

## Find whether an array is a subset of another array using the Frequency Table

The idea is to store the frequency of the elements present in the first array, then look for the elements present in arr2[] in the frequency array. As no new elements are found in arr2[] hence is the subset.

Illustration:

Given array arr1[] = { 11, 1, 13, 21, 3, 7 } and arr2[] = { 11, 3, 7, 1 }.

Step 1: We will store the array arr1[] elements frequency in the frequency array

• The frequency array will look like this

frequency array

Step 2: We will look for arr2[] elements in the frequency array.

• arr2[] = { 11, 3, 7, 1 }, 11 is present in the frequency array
• arr2[] = { 11, 3, 7, 1 }, 3 is present in the frequency array
• arr2[] = { 11, 3, 7, 1 }, 7 is present in the frequency array
• arr2[] = { 11, 3, 7, 1 }, 1 is present in the frequency array

As all the elements are found we can conclude arr2[] is the subset of arr1[].

Algorithm:

The algorithm is pretty straightforward.

• Store the frequency of the first array elements of arr1[] in the frequency array.
• Iterate on the arr2[] and look for its elements in the frequency array.
• If the value is found in the frequency array reduce the frequency value by one.
• If for any elements in arr2[] frequency is less than 1, we will conclude arr2[] is not the subset of arr1[],

Below is the implementation of the above approach:

## C++14

 // C++ program to find whether an array // is subset of another array #include using namespace std; /* Return true if arr2[] is a subset of arr1[] */   bool isSubset(int arr1[], int m, int arr2[], int n) {     // Create a Frequency Table using STL     map frequency;       // Increase the frequency of each element     // in the frequency table.     for (int i = 0; i < m; i++) {         frequency[arr1[i]]++;     }     // Decrease the frequency if the     // element was found in the frequency     // table with the frequency more than 0.     // else return 0 and if loop is     // completed return 1.     for (int i = 0; i < n; i++) {         if (frequency[arr2[i]] > 0)             frequency[arr2[i]]--;         else {             return false;         }     }     return 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, m, arr2, n))         cout << "arr2[] is subset of arr1[] "              << "\n";     else         cout << "arr2[] is not a subset of arr1[] "              << "\n";     return 0; } // This code is contributed by Dhawal Sarin.

## Java

 // Java program to find whether an array // is subset of another array import java.io.*; import java.util.*;   class GFG {       // Return true if arr2[] is a subset of arr1[]     static boolean isSubset(int[] arr1, int m, int[] arr2,                             int n)     {           // Create a Frequency Table using STL         HashMap frequency             = new HashMap();           // Increase the frequency of each element         // in the frequency table.         for (int i = 0; i < m; i++) {             frequency.put(arr1[i],                           frequency.getOrDefault(arr1[i], 0)                               + 1);         }           // Decrease the frequency if the         // element was found in the frequency         // table with the frequency more than 0.         // else return 0 and if loop is         // completed return 1.         for (int i = 0; i < n; i++) {             if (frequency.getOrDefault(arr2[i], 0) > 0)                 frequency.put(arr2[i],                               frequency.get(arr1[i]) - 1);             else {                 return false;             }         }         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, m, arr2, n))             System.out.println(                 "arr2[] is subset of arr1[] ");         else             System.out.println(                 "arr2[] is not a subset of arr1[] ");     } }   // This code is contributed by akhilsaini

## Python3

 # Python3 program to find whether an array # is subset of another array   # Return true if arr2[] is a subset of arr1[]     def isSubset(arr1, m, arr2, n):       # Create a Frequency Table using STL     frequency = {}       # Increase the frequency of each element     # in the frequency table.     for i in range(0, m):         if arr1[i] in frequency:             frequency[arr1[i]] = frequency[arr1[i]] + 1         else:             frequency[arr1[i]] = 1       # Decrease the frequency if the     # element was found in the frequency     # table with the frequency more than 0.     # else return 0 and if loop is     # completed return 1.     for i in range(0, n):         if (frequency[arr2[i]] > 0):             frequency[arr2[i]] -= 1         else:             return False       return True     # 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, m, arr2, n)):         print("arr2[] is subset of arr1[] ")     else:         print("arr2[] is not a subset of arr1[] ")   # This code is contributed by akhilsaini

## C#

 // C# program to find whether an array // is subset of another array using System; using System.Collections; using System.Collections.Generic;   class GFG {       // Return true if arr2[] is a subset of arr1[]     static bool isSubset(int[] arr1, int m, int[] arr2,                          int n)     {           // Create a Frequency Table using STL         Dictionary frequency             = new Dictionary();           // Increase the frequency of each element         // in the frequency table.         for (int i = 0; i < m; i++) {             if (frequency.ContainsKey(arr1[i]))                 frequency[arr1[i]] += 1;             else                 frequency[arr1[i]] = 1;         }           // Decrease the frequency if the         // element was found in the frequency         // table with the frequency more than 0.         // else return 0 and if loop is         // completed return 1.         for (int i = 0; i < n; i++) {             if (frequency[arr2[i]] > 0)                 frequency[arr2[i]] -= 1;             else {                 return false;             }         }         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, m, arr2, n))             Console.WriteLine(                 "arr2[] is subset of arr1[] ");         else             Console.WriteLine(                 "arr2[] is not a subset of arr1[] ");     } }   // This code is contributed by akhilsaini

## Javascript



Output

arr2[] is subset of arr1[]

Time Complexity: O(m+n) which is better than methods 1,2,3
Auxiliary Space: O(n)

Note that method 1, method 2, method 4, and method 5 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.

My Personal Notes arrow_drop_up
Related Articles