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

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  • Last Updated : 25 May, 2021
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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 : O(n*log(n))
 


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