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 } 

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 <bits/stdc++.h>
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[0]);
    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

Javascript

<script>

    // Javascript implementation of the approach

    // Function to count minimum number of subarrays
    function minimumSubarrays(ar, n)
    {
        let se = new Set();

        let cnt = 1;

        for (let i = 0; i < n; i++)
        {

            // Checking if an element already exist in
            // the current sub-array
            if (se.has(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

        let ar = [ 1, 2, 1, 3, 4, 2, 4, 4, 4 ];
        let n = ar.length ;

        document.write(minimumSubarrays(ar, n));

// This code is contributed by susmitakundugoaldanga.
</script>

Output

5

Time Complexity : 
Auxiliary Space: O(n)

Method 2 Using Unordered_map

instead of using set we can use unordered_map for counting subarray without repeating elements in them. as above discussed if we use set in solution then it takes O(n*log(n)) time complexity rather than that if we use unordered_map in solution then time complexity will be O(n). so it’s better in terms of time complexity. below is the code for this solution

C++

// C++ program to count minimum subarray having unique elements

#include <bits/stdc++.h>
using namespace std;

// Function to count minimum number of subarrays
int minimumSubarrays(int ar[], int n)
{
    // map to store the element
    unordered_map<int, int> mp;

    int cnt = 1;

    for (int i = 0; i < n; i++) {
        // Checking if an element already exist in the current sub-array
        if (mp[ar[i]]) {
            cnt++;
            // clear map for new possible value of subarrays
            mp.clear();
            // inserting the current element
            mp[ar[i]]++;
        }
        else {
            // inserting the current element
            mp[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[0]);
      // function call
    cout << minimumSubarrays(ar, n);
    return 0;
}

Java

// Java program to count minimum subarray having unique elements
import java.util.*;

class GFG {

  // Function to count minimum number of subarrays
  public static int minimumSubarrays(int ar[], int n)
  {

    // map to store the element
    Map < Integer, Integer > mp = new HashMap < Integer, Integer > ();

    int cnt = 1;

    for (int i = 0; i < n; i++) {
      // Checking if an element already exist in the current sub-array
      if (mp.containsKey(ar[i])) {
        cnt++;
        // clear map for new possible value of subarrays
        mp.clear();
        // inserting the current element
        mp.put(ar[i], 1);
      } else {
        // inserting the current element
        mp.put(ar[i], 1);
      }
    }

    return cnt;
  }

  public static void main(String[] args) {
    int ar[] = {1, 2, 1, 3, 4, 2, 4, 4, 4 };
    int n = ar.length;

    // function call
    System.out.println(minimumSubarrays(ar, n));
  }
}

// This code is contributed by ajaymakavana.

Python3

# Python program to count minimum subarray
# having unique elements

# function to count minimum number of subarrays
def minimumSubarrays(ar, n):

    # map to store the element
    mp = {}

    cnt = 1

    for i in range(n):

        # checking if an element already
        # exist in the current sub-array
        if(ar[i] in mp):
            cnt += 1

            # clear map for new possible
            # value of subarrays
            mp.clear()

            # Inserting the current element
            mp[ar[i]] = 1
        else:

            # Inserting the current element
            mp[ar[i]] = 1

    return cnt


ar = [1, 2, 1, 3, 4, 2, 4, 4, 4]
n = len(ar)

# Function call
print(minimumSubarrays(ar, n))

# This code is contributed by lokesh.

C#

// C# program to count minimum subarray having unique elements
using System;
using System.Collections.Generic;

public class GFG {

  // Function to count minimum number of subarrays
  public static int minimumSubarrays(int[] ar, int n) {

    // map to store the element
    Dictionary < int, int > mp = new Dictionary < int, int > ();

    int cnt = 1;

    for (int i = 0; i < n; i++) {
      // Checking if an element already exist in the current sub-array
      if (mp.ContainsKey(ar[i])) {
        cnt++;
        // clear map for new possible value of subarrays
        mp.Clear();
        // inserting the current element
        mp[ar[i]] = 1;
      } else {
        // inserting the current element
        mp[ar[i]] = 1;
      }
    }

    return cnt;
  }

  public static void Main(string[] args) {
    int[] ar = {1, 2, 1, 3, 4, 2, 4, 4, 4 };
    int n = ar.Length;

    // function call
    Console.WriteLine(minimumSubarrays(ar, n));
  }
}

// This code is contributed by ajaymakvana

Javascript

// Javascript program to count minimum subarray having unique elements

// Function to count minimum number of subarrays
function minimumSubarrays(ar,n)
{
    // map to store the element
    const mp = new Map();
    let cnt = 1;

    for (let i = 0; i < n; i++)
    {

        // Checking if an element already exist in the current sub-array
        if (mp.has(ar[i]) == true)
        {
            cnt++;

            // clear map for new possible value of subarrays
            mp.clear();

            // inserting the current element
            mp.set(ar[i],1);
        }
        else
        {
            // inserting the current element
            mp.set(ar[i],1);
        }
    }

    return cnt;
}

// Driver Code

    let ar = [1, 2, 1, 3, 4, 2, 4, 4, 4 ];
    let n = ar.length;

      // function call
    console.log(minimumSubarrays(ar, n));

   // This code is contributed by garg28harsh.

Output

5

Time Complexity: O(n)

Auxiliary Space: O(n)