# Java Programming – Find whether an array is subset of another array

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.

Given two arrays: arr1[0..m-1] and arr2[0..n-1]. Find whether arr2[] is a subset of arr1[] or not. Both the arrays are not in sorted order. It may be assumed that elements in both array are distinct.

Examples:
Input: arr1[] = {11, 1, 13, 21, 3, 7}, arr2[] = {11, 3, 7, 1}
Output: arr2[] is a subset of arr1[]

Input: arr1[] = {1, 2, 3, 4, 5, 6}, arr2[] = {1, 2, 4}
Output: arr2[] is a subset of arr1[]

Input: arr1[] = {10, 5, 2, 23, 19}, arr2[] = {19, 5, 3}
Output: arr2[] is not a subset of arr1[]

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

Java Program
``````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;
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;
}

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[]");
}
}``````

Output:

```arr2[] is subset of arr1[]
```

Time Complexity: O(m*n)

Method 2 (Use Sorting and Binary Search)

```1) Sort arr1[] O(mLogm)
2) For each element of arr2[], do binary search for it in sorted arr1[].
3) If all elements are present then return 1.```
Java Program
``````class Main
{
/* Return true if arr2[] is a subset of arr1[] */
static boolean isSubset(int arr1[], int arr2[], int m, int n)
{
int i = 0;

sort(arr1, 0, m-1);
for (i=0; i<n; i++)
{
if (binarySearch(arr1, 0, m-1, arr2[i]) == -1)
return false;
}

/* If we reach here then all elements of arr2[]
are present in arr1[] */
return true;
}

/* FOLLOWING FUNCTIONS ARE ONLY FOR SEARCHING AND SORTING PURPOSE */
/* Standard Binary Search function*/
static int binarySearch(int arr[], int low, int high, int x)
{
if(high >= low)
{
int mid = (low + high)/2;  /*low + (high - low)/2;*/

/* Check if arr[mid] is the first occurrence of x.
arr[mid] is first occurrence if x is one of the following
is true:
(i)  mid == 0 and arr[mid] == x
(ii) arr[mid-1] < x and arr[mid] == x
*/
if(( mid == 0 || x > arr[mid-1]) && (arr[mid] == x))
return mid;
else if(x > arr[mid])
return binarySearch(arr, (mid + 1), high, x);
else
return binarySearch(arr, low, (mid -1), x);
}
return -1;
}

/* This function takes last element as pivot,
places the pivot element at its correct
position in sorted array, and places all
smaller (smaller than pivot) to left of
pivot and all greater elements to right
of pivot */
static int partition(int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low-1); // index of smaller element
for (int j=low; j<high; j++)
{
// If current element is smaller than or
// equal to pivot
if (arr[j] <= pivot)
{
i++;

// swap arr[i] and arr[j]
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}

// swap arr[i+1] and arr[high] (or pivot)
int temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;

return i+1;
}

/* The main function that implements QuickSort()
arr[] --> Array to be sorted,
low  --> Starting index,
high  --> Ending index */
static void sort(int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[pi] is
now at right place */
int pi = partition(arr, low, high);

// Recursively sort elements before
// partition and after partition
sort(arr, low, pi-1);
sort(arr, pi+1, high);
}
}

public static void main(String args[])
{
int arr1[] = {11, 1, 13, 21, 3, 7};
int arr2[] = {11, 3, 7, 1};

int m = arr1.length;
int n = arr2.length;

if(isSubset(arr1, arr2, m, n))
System.out.print("arr2[] is subset of arr1[] ");
else
System.out.print("arr2[] is not a subset of arr1[]");
}
}``````

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 sued and worst case time complexity of Quick Sort is O(m^2) 