# Divide the array into minimum number of sub-arrays having unique elements

Given an array arr. The task is to divide the array into the minimum number of subarrays containing unique elements and return the count of such subarrays.

Note: An array element cannot be present in more than one subarray.

Examples :

```Input : arr[] = {1, 2, 1, 1, 2, 3}
Output : 3
Explanation : The subarrays having unique elements are
{ 1, 2 }, { 1 }, and { 1, 2, 3 }

Input : arr[] = {1, 2, 3, 4, 5}
Output : 1
Explanation : The subarray having unique elements is
{ 1, 2, 3, 4, 5 }
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The idea is to maintain a set while traversing the array. While traversing, if an element is already found in the set, then increase the count of subarray by 1 as we have to include the current element in the next subarray and clear the set for new subarray. Then, proceed for the complete array in a self-similar manner. The variable storing the count will be the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to count minimum subarray having ` `// unique elements ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count minimum number of subarrays ` `int` `minimumSubarrays(``int` `ar[], ``int` `n) ` `{ ` `    ``set<``int``> se; ` ` `  `    ``int` `cnt = 1; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// Checking if an element already exist in  ` `        ``// the current sub-array ` `        ``if` `(se.count(ar[i]) == 0) { ` `            ``// inserting the current element ` `            ``se.insert(ar[i]); ` `        ``} ` `        ``else` `{ ` `            ``cnt++; ` `            ``// clear set for new possible value of subarrays ` `            ``se.clear(); ` `            ``// inserting the current element ` `            ``se.insert(ar[i]); ` `        ``} ` `    ``} ` ` `  `    ``return` `cnt; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `ar[] = { 1, 2, 1, 3, 4, 2, 4, 4, 4 }; ` `    ``int` `n = ``sizeof``(ar) / ``sizeof``(ar); ` `    ``cout << minimumSubarrays(ar, n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to count minimum number of subarrays  ` `    ``static` `int` `minimumSubarrays(``int` `ar[], ``int` `n)  ` `    ``{  ` `        ``Vector se = ``new` `Vector(); ` `     `  `        ``int` `cnt = ``1``;  ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{  ` `             `  `            ``// Checking if an element already exist in  ` `            ``// the current sub-array  ` `            ``if` `(se.contains(ar[i]) == ``false``)  ` `            ``{  ` `                ``// inserting the current element  ` `                ``se.add(ar[i]);  ` `            ``}  ` `            ``else` `            ``{  ` `                ``cnt++;  ` `                 `  `                ``// clear set for new possible value  ` `                ``// of subarrays  ` `                ``se.clear(); ` `                 `  `                ``// inserting the current element  ` `                ``se.add(ar[i]);  ` `            ``}  ` `        ``}  ` `        ``return` `cnt;  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `ar[] = { ``1``, ``2``, ``1``, ``3``, ``4``, ``2``, ``4``, ``4``, ``4` `};  ` `        ``int` `n = ar.length ; ` `         `  `        ``System.out.println(minimumSubarrays(ar, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python 3 implementation of the approach  ` ` `  `# Function to count minimum number of subarrays  ` `def` `minimumSubarrays(ar, n) :  ` `    ``se ``=` `[]  ` ` `  `    ``cnt ``=` `1``;  ` ` `  `    ``for` `i ``in` `range``(n) : ` `         `  `        ``# Checking if an element already exist in  ` `        ``# the current sub-array  ` `        ``if` `se.count(ar[i]) ``=``=` `0` `:  ` `             `  `            ``# inserting the current element  ` `            ``se.append(ar[i])  ` `        ``else` `:  ` `            ``cnt ``+``=` `1` `             `  `            ``# clear set for new possible value  ` `            ``# of subarrays  ` `            ``se.clear()  ` `             `  `            ``# inserting the current element  ` `            ``se.append(ar[i])  ` `    ``return` `cnt  ` ` `  `# Driver Code  ` `ar ``=` `[ ``1``, ``2``, ``1``, ``3``, ``4``, ``2``, ``4``, ``4``, ``4` `]  ` `n ``=` `len``(ar)  ` `print``(minimumSubarrays(ar, n))  ` ` `  `# This code is contributed by ` `# divyamohan123 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic;              ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to count minimum number of subarrays  ` `    ``static` `int` `minimumSubarrays(``int` `[]ar, ``int` `n)  ` `    ``{  ` `        ``List<``int``> se = ``new` `List<``int``>(); ` `     `  `        ``int` `cnt = 1;  ` `     `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{  ` `             `  `            ``// Checking if an element already exist in  ` `            ``// the current sub-array  ` `            ``if` `(se.Contains(ar[i]) == ``false``)  ` `            ``{  ` `                ``// inserting the current element  ` `                ``se.Add(ar[i]);  ` `            ``}  ` `            ``else` `            ``{  ` `                ``cnt++;  ` `                 `  `                ``// clear set for new possible value  ` `                ``// of subarrays  ` `                ``se.Clear(); ` `                 `  `                ``// inserting the current element  ` `                ``se.Add(ar[i]);  ` `            ``}  ` `        ``}  ` `        ``return` `cnt;  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{  ` `        ``int` `[]ar = { 1, 2, 1, 3, 4, 2, 4, 4, 4 };  ` `        ``int` `n = ar.Length ; ` `         `  `        ``Console.WriteLine(minimumSubarrays(ar, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```5
```

Time Complexity : My Personal Notes arrow_drop_up Computer Science Student || Competitive Coder || Machine Learning Enthusiast || Open to internship opportunities

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