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# Find whether an array is subset of another array

The problem “Find whether an array is subset of another array” states that you are given two arrays arra1[] and array2[]. The arrays given are in an unsorted manner. Your task is to find whether the array2[] is a subset of array1[].

## Example ```arr1= [1,4,5,7,8,2]
arr2= [1,7,2,4]```
`arr2 [] is a subset of arr1 [].`
```arr1= [21,4,5,2,18,3]
arr2= [1,4,2,3]```
`arr2 [] is not a subset of arr1 [].`

## Algorithm

1. Open a nested loop.
2. Set i to 0, j to 0.
3. While i is less than the length of array2[].
1. While j is less than the length of array1[].
1. If arr2[i] is equal to arr[j], then break.
4. If j is equal to m, return false.
5. Return true.

### Explanation

Our task is to find whether the array second is a subset of array1. So our main idea is to use the nested loop and check each element and array2’s every element is found in array1, which means array2 is a subset of array1.

Let us take an example as our input in array1 and array2

Example

arr1[] = {11, 1, 13, 21, 3, 7}; arr2[] = {11, 3, 7, 1};

Starting with the 0th index of array2, we are going to check if the 0th index of arrays2 finds a similar number in array1[i], and yes it found it as 11. So we are going to break the loop and do i++.

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Starting with the 1st index of array2[], we are going to check if the 1st index of arrays2 finds a similar number in array1[i], and yes it found it as 3. So we are going to break the loop and do i++.

Starting with the 2nd index of array2[], we are going to check if the 2nd index of arrays2 finds a similar number in array1[i] and it found it as 7. So we are going to break the loop and do i++.

Starting with the 1st index of array2[], we are going to check if the 1st index of arrays2 finds a similar number in array1[i] and 1 found it. So we are going to break the loop and do i++.

And about if condition which is represented as if ( j = = m ) this means, if in any of the iteration if any of array2’s element is not found in array1[], means it comes out of the loop without breaking it means ‘j’ with a value equal to the length of the array1[] that means we didn’t find one of the elements similar in array1[] and we return false because subset consists all of the element of the given set and we didn’t find one.

## Code

### C++ code to find whether an array is subset of another array

```#include <iostream>
using namespace std;

bool isSubset(int arr1[], int arr2[], int l1, int l2)
{
int i,j;
for(i=0; i<l2; i++)
{

for(j=0; j<l1; j++)
{
if(arr2[i]==arr1[j])
break;
}
if(j==l1)
return false;
}
return true;
}
int main()
{
int arr1[] = {1, 2, 3, 4, 5, 6};
int arr2[] = {1, 3, 5};

int l1=sizeof(arr1)/sizeof(arr1);
int l2=sizeof(arr2)/sizeof(arr2);
if(isSubset(arr1,arr2,l1,l2))
{
cout<<"arr2[] is a subset of arr1[]";
}
else
{
cout <<"arr2[] is not a subset of arr1[]"<<endl;
}
return 0;
}
```
`arr2[] is a subset of arr1[]`

### Java code to find whether an array is subset of another array

```class isArraySubset
{
public static boolean isSubset(int arr1[],int arr2[], int l1, int l2)
{
int i = 0;
int j = 0;
for (i = 0; i < l2; i++)
{
for (j = 0; j < l1; j++)
{
if(arr2[i] == arr1[j])
{
break;
}
}
if (j == l1)
return false;
}
return true;
}
public static void main(String args[])
{
int arr1[] = {1, 2, 3, 4, 5, 6};
int arr2[] = {1, 3, 5};

int l1 = arr1.length;
int l2 = arr2.length;

if(isSubset(arr1, arr2, l1, l2))
{
System.out.println("arr2[] is a subset of arr1[]");
}
else
{
System.out.println("arr2[] is not a subset of arr1[]");
}
}
}
```
`arr2[] is a subset of arr1[]`

## Complexity Analysis

### Time Complexity

O(m*n) where “m” is the size of arr1 and “n” is the size of arr2. Because we have used nested loops, making the time complexity polynomial.

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### Space Complexity

O(1), because we have not used any extra array/vector. 