Maximum distinct elements after removing k elements

Given an array arr[] containing n elements. The problem is to find maximum number of distinct elements (non-repeating) after removing k elements from the array.
Note: 1 <= k <= n.

Examples:

Input : arr[] = {5, 7, 5, 5, 1, 2, 2}, k = 3
Output : 4
Remove 2 occurrences of element 5 and
1 occurrence of element 2.

Input : arr[] = {1, 2, 3, 4, 5, 6, 7}, k = 5
Output : 2

Input : arr[] = {1, 2, 2, 2}, k = 1
Output : 1



Approach: Following are the steps:

  1. Create a hash table to store the frequency of each element.
  2. Insert frequency of each element in a max heap.
  3. Now, perform the following operation k times. Remove an element from the max heap. Decrement its value by 1. After this if element is not equal to 0, then again push the element in the max heap.
  4. After the completion of step 3, the number of elements in the max heap is the required answer.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to find maximum distinct 
// elements after removing k elements
#include <bits/stdc++.h>
using namespace std;
  
// function to find maximum distinct elements
// after removing k elements
int maxDistinctNum(int arr[], int n, int k)
{
    // 'um' implemented as hash table to store
    // frequency of each element
    unordered_map<int, int> um;
  
    // priority_queue 'pq' implemented as
    // max heap
    priority_queue<int> pq;
  
    // storing frequency of each element in 'um'
    for (int i = 0; i < n; i++)
        um[arr[i]]++;
  
    // inserting frequency of each element in 'pq'
    for (auto it = um.begin(); it != um.end(); it++)
        pq.push(it->second);
  
    while (k--) {
  
        // get the top element of 'pq'
        int temp = pq.top();
  
        // remove top element from 'pq'
        pq.pop();
  
        // decrement the popped element by 1
        temp--;
  
        // if true, then push the element in 'pq'
        if (temp != 0)
            pq.push(temp);
    }
  
    // Count all those elements that appear
    // once after above operations.
    int res = 0;
    while (pq.empty() == false)
    {
        int x = pq.top();
        pq.pop();
        if (x == 1)
          res++;
    }      
  
    return res;
}
  
// Driver program to test above
int main()
{
    int arr[] = { 5, 7, 5, 5, 1, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    cout << "Maximum distinct elements = "
         << maxDistinctNum(arr, n, k);
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java implementation to find maximum distinct   


// elements after removing k elements 


import java.util.*; 


  


class GFG { 


      // function to find maximum distinct elements  


      // after removing k elements  


      static int maxDistinctNum(int[] arr, int n, int k) 


      { 


             // hash map to store  


             // frequency of each element  


             HashMap<Integer, Integer> map = new HashMap<>(); 


  


             // priority_queue 'pq' implemented as  


             // max heap  


             PriorityQueue<Integer> pq =  


                         new PriorityQueue<>(Collections.reverseOrder()); 


               


             // storing frequency of each element in map 


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


                  if(map.containsKey(arr[i])) { 


                       int val = map.get(arr[i]); 


                       val++; 


                       map.remove(arr[i]); 


                       map.put(arr[i], val); 


                    } 


  


                  else  


                      map.put(arr[i], 1); 


             } 


  


             // inserting frequency of each element in 'pq' 


             for (Map.Entry<Integer, Integer> entry : map.entrySet()) { 


                  pq.add(entry.getValue()); 


             } 


  


             while (k > 0) { 


                   // get the top element of 'pq' 


                   int temp = pq.poll(); 


  


                   // decrement the popped element by 1  


                   temp--;  


  


                   // if true, then push the element in 'pq' 


                   if (temp > 0) 


                       pq.add(temp); 


                   k--; 


             }  


  


             // Count all those elements that appear  


             // once after above operations.  


             int res = 0; 


             while (pq.size() != 0) { 


                   pq.poll(); 


                   res++; 


             } 


  


             return res; 


      } 


  


      // Driver code 


      public static void main(String args[]) 


      { 


             int[] arr = { 5, 7, 5, 5, 1, 2, 2 }; 


             int n = arr.length; 


             int k = 3; 


             System.out.println("Maximum distinct elements = " +  


                                maxDistinctNum(arr, n, k)); 


      } 


}  


  


// This code is contributed by rachana soma 



















                        chevron_right







                        filter_none










Output:


Maximum distinct elements = 4



Time Complexity: O(k*logd), where d is the number of distinct elements in the given array.



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
ayushjauhari14
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.


Improved By :  rachana soma