# 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++

 `// 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); ` `    ``int` `n = ``sizeof``(arr2)/``sizeof``(arr2); ` ` `  `    ``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[]"``);  ` `    ``} ` `} `

## Python 3

 `# 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 `

## PHP

 ` `

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++

 `// 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); ` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2); ` ` `  `    ``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 ` `/* 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); ` `    ``int` `n = ``sizeof``(arr2)/``sizeof``(arr2); ` ` `  `    ``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= 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

 `# 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= 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 `

## PHP

 `= ``\$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++

 `// 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); ` `    ``int` `n = ``sizeof``(arr2)/``sizeof``(arr2); ` ` `  `    ``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``; ` `    ``}  ` `         `  `    ``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

 `# 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 a 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 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

 `// 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#

 `// 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 `

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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