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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

5

Time Complexity :  O(n*log(n))



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