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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++

// 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;
}

Java

// 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


Output:

Maximum distinct elements = 4

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



This article is attributed to GeeksforGeeks.org

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