# Maximum number of Unique integers in Sub-Array of given size

Given an array of N integers and a number M. The task is to find out the maximum number of unique integers among all possible contiguous subarrays of size M.

Examples:

Input : arr[] = {5, 3, 5, 2, 3, 2}, M = 3
Output : 3
Explanation:
In the sample test case, there are 4 subarrays of size 3.
s1 = (5, 3, 5)- Has 2 unique numbers.
s2 = (3, 5, 2)- Has 3 unique numbers.
s3 = (5, 2, 3)- Has 3 unique numbers.
s4 = (2, 3, 2)- Has 2 unique numbers.
In these subarrays, there are 2, 3, 3, 2 unique numbers, respectively.
The maximum amount of unique numbers among all possible contiguous subarrays is 3.

Input : arr[] = {5, 5, 5, 5, 5, 5}, M = 3
Output : 1

Naive Approach:

1. Generate all subarrays of size M.
2. Count unique number for each subarray.
• Check whether it is greater than the previous maximum unique number or not, if yes, replace it with the previous maximum unique number.
3. Continue until we generate all possible subarrays.

Below is the implementation of the above approach:

## C++

 `// A C++ programme to find maximum distinct elements ` `// in a subarray of size k ` `#include ` `using` `namespace` `std; ` `//Function to find maximun unique element in  ` `//a subarray of size k ` `int` `maxUniqueNum(``int` `a[],``int` `N,``int` `M) ` `{ ` `    ``int` `maxUnique=0; ` `    ``//search every subarray of size k  ` `    ``//and find how many unique element present ` `    ``for``(``int` `i=0;i s; ` `        ``for``(``int` `j=0;jmaxUnique) ` `        ``{ ` `            ``maxUnique=s.size(); ` `        ``} ` `    ``} ` `    ``return` `maxUnique; ` `} ` `     `  `int` `main() ` `{ ` `    ``int` `arr[] = {5, 3, 5, 2, 3, 2}; ` `    ``int` `M=3,N=``sizeof``(arr)/``sizeof``(arr[0]); ` `    ``cout<

## Java

 `// Java Program to find maximum number of ` `// Unique integers in Sub-Array ` `// of given size ` ` `  `import` `java.util.*; ` `class` `GFG { ` ` `  `    ``// Function to find maximum number of ` `    ``// Unique integers in Sub-Array ` `    ``// of given size ` `    ``public` `static` `int` `maxUniqueNum(``int` `arr[], ` `                                   ``int` `N, ``int` `M) ` `    ``{ ` `        ``int` `maxUnique = ``0``; ` ` `  `        ``// Generate all subarrays of size M ` `        ``for` `(``int` `i = ``0``; i < N - M; i++) { ` `            ``int` `currentUnique = ``0``; ` ` `  `            ``HashMap map = ``new` `HashMap(); ` ` `  `            ``for` `(``int` `k = i; k < i + M; k++) { ` ` `  `                ``// if the key is new to the map, ` `                ``// push the key in map and increment ` `                ``// count for unique number ` `                ``if` `(!map.containsKey(arr[k])) { ` `                    ``map.put(arr[i], ``1``); ` `                    ``currentUnique++; ` `                    ``continue``; ` `                ``} ` `            ``} ` ` `  `            ``if` `(currentUnique > maxUnique) ` `                ``maxUnique = currentUnique; ` `        ``} ` ` `  `        ``return` `maxUnique; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] arr = { ``5``, ``3``, ``5``, ``2``, ``3``, ``2` `}; ` `        ``int` `N = ``6``; ` ` `  `        ``int` `M = ``3``; ` ` `  `        ``System.out.println(maxUniqueNum(arr, N, M)); ` `    ``} ` `} `

## Python3

# A python 3 programme to find maximum
# distinct elements in a subarray of size k

# Function to find maximun unique
# element in a subarray of size k
def maxUniqueNum(a, N, M):
maxUnique = 0

# search every subarray of size k and
# find how many unique element present
for i in range(N – M):

# create an empty set to store
# the unique elements
s = set()

for j in range(M):
# insert all elements
# duplicate elements are not
# stored in set
s.add(a[i + j])

# update the maxUnique
if(len(s) > maxUnique):
maxUnique = len(s)

return maxUnique

# Driver Code
if __name__ == ‘__main__’:
arr = [5, 3, 5, 2, 3, 2]
M = 3
N = len(arr)
print(maxUniqueNum(arr, N, M))

# This code is contributed by
# Sanjit_Prasad

Output:

```3
```

Time Complexity : O(M * N)
Auxiliary Space : O(M)

Efficient Solution An efficient solution is to use window sliding technique. We maintain a single hash table for storing unique elements of every window.
1) Store counts of first M elements in a hash map.
2) Traverse from (M+1)-th element and for every element, add it to hash map and remove first element of previous window.

Below is the implementation of the above approach:

## C++

 `// An efficient Approach to count distinct elements in ` `// every window of size k ` `#include ` `using` `namespace` `std; ` `//Function to find maximum unique element in ` `//a subarray of size k ` `int` `max_U_element(``int` `a[],``int` `N,``int` `M) ` `{ ` `    ``//map to store the unique elements and their size ` `    ``map<``int``,``int``> hash; ` `    ``//Number of unique elements in an window ` `    ``int` `dist_count=0;      ` `    ``int` `res=0;     ``//Maximum unique element in a window ` `    ``//store all elements till size k i.e. ` `    ``//storing first window ` `    ``for``(``int` `i=0;i

## Java

 `// An efficient Java program to count distinct elements in ` `// every window of size k ` `import` `java.util.HashMap; ` ` `  `class` `maxUniqueNumWindow { ` `    ``static` `int` `maxUniqueNum(``int` `arr[], ``int` `M) ` `    ``{ ` `        ``// Creates an empty hashMap hM ` `        ``HashMap hM = ``new` `HashMap(); ` ` `  `        ``// initialize distinct element count for ` `        ``// current window ` `        ``int` `dist_count = ``0``; ` ` `  `        ``// Traverse the first window and store count ` `        ``// of every element in hash map ` `        ``for` `(``int` `i = ``0``; i < M; i++) { ` `            ``if` `(hM.get(arr[i]) == ``null``) { ` `                ``hM.put(arr[i], ``1``); ` `                ``dist_count++; ` `            ``} ` `            ``else` `{ ` `                ``int` `count = hM.get(arr[i]); ` `                ``hM.put(arr[i], count + ``1``); ` `            ``} ` `        ``} ` `  `  `        ``int` `res = dist_count; ` ` `  `        ``// Traverse through the remaining array ` `        ``for` `(``int` `i = M; i < arr.length; i++) { ` ` `  `            ``// Remove first element of previous window ` `            ``// If there was only one occurrence, then ` `            ``// reduce distinct count. ` `            ``if` `(hM.get(arr[i - M]) == ``1``) { ` `                ``hM.remove(arr[i - M]); ` `                ``dist_count--; ` `            ``} ` ` `  `            ``else` `// reduce count of the removed element ` `            ``{ ` `                ``int` `count = hM.get(arr[i - M]); ` `                ``hM.put(arr[i - M], count - ``1``); ` `            ``} ` ` `  `            ``// Add new element of current window ` `            ``// If this element appears first time, ` `            ``// increment distinct element count ` `            ``if` `(hM.get(arr[i]) == ``null``) { ` `                ``hM.put(arr[i], ``1``); ` `                ``dist_count++; ` `            ``} ` `            ``else` `// Increment distinct element count ` `            ``{ ` `                ``int` `count = hM.get(arr[i]); ` `                ``hM.put(arr[i], count + ``1``); ` `            ``} ` ` `  `            ``res = Math.max(res, dist_count); ` `        ``} ` ` `  `        ``return` `res; ` `    ``} ` ` `  `    ``// Driver method ` `    ``public` `static` `void` `main(String arg[]) ` `    ``{ ` `        ``int` `arr[] = { ``1``, ``2``, ``1``, ``3``, ``4``, ``2``, ``3` `}; ` `        ``int` `M = ``4``; ` `        ``System.out.println(maxUniqueNum(arr, M)); ` `    ``} ` `} `

Output:

```4
```

Time Complexity : O(N)
Auxiliary Space : O(M)

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