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Remove an occurrence of most frequent array element exactly K times

Last Updated : 15 Jun, 2021
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Given an array arr[], the task is to remove an occurrence of the most frequent array element exactly K times. If multiple array elements have maximum frequency, remove the smallest of them. Print the K deleted elements.

Examples:

Input: arr[] = {1, 3, 2, 1, 4, 1}, K = 2
Output: 1 1
Explanation: 
The frequency of 1 is 3 and frequencies of 2, 3, 4 are 1:
Operation 1: Remove 1 from the array. Currently, the frequency of 1 is 2 and frequencies of 2, 3, 4 is 1.
Operation 2: Remove 1 from the array.

Input: arr[] = {10, 10, 10, 20, 30, 20, 20}, K = 2
Output: 10 20

Naive Approach: The simplest approach is to sort the array in ascending order and count the frequencies of array elements using a Map. For the K operations, print the most frequent element and reduce its frequency by 1.

Below is the implementation of the above approach:

C++ 
// C++ program for the above approach

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

// Function to print the most frequent
// array element exactly K times
void maxFreqElements(int arr[],
                     int N, int K)
{
    // Stores frequency array element
    map<int, int> mp;

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

        // Count frequency
        // of array element
        mp[arr[i]]++;
    }

    while (K > 0) {

        // Maximum array element
        int max = 0;
        int element;

        // Traverse the Map
        for (auto i : mp) {

            // Find the element with
            // maximum frequency
            if (i.second > max) {
                max = i.second;

                // If the frequency is maximum,
                // store that number in element
                element = i.first;
            }
        }

        // Print element as it contains the
        // element having highest frequency
        cout << element << " ";

        // Decrease the frequency
        // of the maximum array element
        mp[element]--;

        // Reduce the number of operations
        K--;
    }
}

// Driver Code
int main()
{

    // Given array
    int arr[] = { 1, 3, 2, 1, 4, 1 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Given K
    int K = 2;

    maxFreqElements(arr, N, K);

    return 0;
}
Java 
// Java program to implement 
// the above approach 
import java.util.*;
class GFG
{

// Function to print the most frequent
// array element exactly K times
static void maxFreqElements(int arr[],
                     int N, int K)
{
  
    // Stores frequency array element
    HashMap<Integer,
          Integer> mp = new HashMap<Integer,
                                    Integer>();
 
    for (int i = 0; i < N; i++) 
    {
 
        // Count frequency
        // of array element
        if(mp.containsKey(arr[i]))
            {
                mp.put(arr[i], mp.get(arr[i]) + 1);
            }
            else
            {
                mp.put(arr[i], 1);
            }
    }
 
    while (K > 0) 
    {
 
        // Maximum array element
        int max = 0;
        int element = 0;
 
        // Traverse the Map
        for (Map.Entry<Integer,
                 Integer> i : mp.entrySet()) 
        {
 
            // Find the element with
            // maximum frequency
            if (i.getValue() > max)
            {
                max = i.getValue();
 
                // If the frequency is maximum,
                // store that number in element
                element = i.getKey();
            }
        }
 
        // Print element as it contains the
        // element having highest frequency
        System.out.print(element + " ");

 
        // Decrease the frequency
        // of the maximum array element
        if(mp.containsKey(element))
            {
                mp.put(element, mp.get(element) + 1);
            }
            else
            {
                mp.put(element, 1);
            }

        // Reduce the number of operations
        K--;
    }
}
 
// Driver code
public static void main(String[] args)
{
    // Given array
    int[] arr = { 1, 3, 2, 1, 4, 1 };
     
    // Size of the array
    int N = arr.length;
     
    // Given K
    int K = 2;   
    maxFreqElements(arr, N, K);
}
}

// This code is contributed by susmitakundugoaldanga
Python3 
# Python3 program for the above approach 

# Function to print the most frequent 
# array element exactly K times 
def maxFreqElements(arr, N, K) : 

    # Stores frequency array element 
    mp = {} 

    for i in range(N) :

        # Count frequency 
        # of array element 
        if arr[i] in mp :
            mp[arr[i]] += 1
        else :
            mp[arr[i]] = 1

    while (K > 0) : 

        # Maximum array element 
        Max = 0

        # Traverse the Map 
        for i in mp :

            # Find the element with 
            # maximum frequency 
            if (mp[i] > Max) : 
                Max = mp[i] 

                # If the frequency is maximum, 
                # store that number in element 
                element = i

        # Print element as it contains the 
        # element having highest frequency 
        print(element, end = " ") 

        # Decrease the frequency 
        # of the maximum array element 
        if element in mp :
            mp[element] -= 1
        else :
            mp[element] = -1

        # Reduce the number of operations 
        K -= 1

# Given array 
arr = [ 1, 3, 2, 1, 4, 1 ] 

# Size of the array 
N = len(arr)  

# Given K 
K = 2

maxFreqElements(arr, N, K)

# This code is contributed by divyeshrabadiya07
C# 
// C# program for the above approach
using System;
using System.Collections.Generic;  

class GFG{
    
// Function to print the most frequent
// array element exactly K times
static void maxFreqElements(int[] arr, 
                            int N, int K)
{
    
    // Stores frequency array element
    Dictionary<int,
               int> mp = new Dictionary<int, 
                                        int>();  
                                        
    for(int i = 0; i < N; i++)
    {
        
        // Count frequency
        // of array element
        if (mp.ContainsKey(arr[i]))
        {
            mp[arr[i]]++;
        }
        else
        {
            mp[arr[i]] = 1;
        }
    }
 
    while (K > 0)
    {
        
        // Maximum array element
        int max = 0;
        int element = 0;
 
        // Traverse the Map
        foreach(KeyValuePair<int, int> i in mp)
        {
            
            // Find the element with
            // maximum frequency
            if (i.Value > max) 
            {
                max = i.Value;
                
                // If the frequency is maximum,
                // store that number in element
                element = i.Key;
            }
        }
 
        // Print element as it contains the
        // element having highest frequency
        Console.Write(element + " ");
 
        // Decrease the frequency
        // of the maximum array element
        if (mp.ContainsKey(element))
        {
            mp[element]--;
        }
        else
        {
            mp[element] = -1;
        }
 
        // Reduce the number of operations
        K--;
    }
}

// Driver Code
static void Main() 
{
    
    // Given array
    int[] arr = { 1, 3, 2, 1, 4, 1 };
    
    // Size of the array
    int N = arr.Length;
    
    // Given K
    int K = 2;
    
    maxFreqElements(arr, N, K);
}
}

// This code is contributed by divyesh072019
JavaScript 
<script>

      // JavaScript program for the above approach
      
      // Function to print the most frequent
      // array element exactly K times
      function maxFreqElements(arr, N, K) {
        // Stores frequency array element
        var mp = {};

        for (var i = 0; i < N; i++) {
          // Count frequency
          // of array element
          if (mp.hasOwnProperty(arr[i])) {
            mp[arr[i]]++;
          } else {
            mp[arr[i]] = 1;
          }
        }

        while (K > 0) {
          // Maximum array element
          var max = 0;
          var element = 0;

          // Traverse the Map
          for (const [key, value] of Object.entries(mp)) {
            // Find the element with
            // maximum frequency
            if (value > max) {
              max = value;

              // If the frequency is maximum,
              // store that number in element
              element = key;
            }
          }

          // Print element as it contains the
          // element having highest frequency
          document.write(element + " ");

          // Decrease the frequency
          // of the maximum array element
          if (mp.hasOwnProperty(element)) {
            mp[element]--;
          } else {
            mp[element] = -1;
          }

          // Reduce the number of operations
          K--;
        }
      }

      // Driver Code
      // Given array
      var arr = [1, 3, 2, 1, 4, 1];

      // Size of the array
      var N = arr.length;

      // Given K
      var K = 2;

      maxFreqElements(arr, N, K);
</script>

Output: 
1 1

 

Time Complexity: O(N * K)
Auxiliary Space: O(N)

Efficient Approach: The idea is to store the array of elements in a vector of pairs along with their count and then sort the vector of pairs in ascending order using a comparator. Once done, print the first K elements from that vector of pairs.

Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++ 
// C++ program for the above approach

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

// Function to sort the vector
// of vector of pair
bool cmp(pair<int, int> p1, pair<int, int> p2)
{
    // Check if frequency of p1 is
    // greater than frequency of p2
    if (p1.second > p2.second)
        return true;

    // If frequency of p1 and p2 is same
    else if (p1.second == p2.second) {

        // Check for the smallest element
        if (p1.first < p2.first)
            return true;
    }
    return false;
}

// Function to print the K most frequent
// elements after each removal
void maxFreqElements(int arr[], int N, int K)
{
    // Stores frequency of array elements
    map<int, int> mp;

    // Pairs array element with frequency
    vector<pair<int, int> > v;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Count the frequencies
        mp[arr[i]]++;

        // Insert the element with its
        // current frequency into the vector
        v.push_back({ arr[i], mp[arr[i]] });
    }

    // Sort the vector according to
    // higher frequency and smaller
    // element if frequency is same
    sort(v.begin(), v.end(), cmp);

    // Print the first K elements
    // of the array
    for (int i = 0; i < K; i++)
        cout << v[i].first << " ";
}

// Driver Code
int main()
{

    // Given array
    int arr[] = { 1, 3, 2, 1, 4, 1 };

    // Given K
    int K = 2;

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    maxFreqElements(arr, N, K);

    return 0;
}
Java 
// Java program for above approach
import java.util.*;
import java.lang.*;

class pair{
    int first,second;
    
    pair(int first, int second){
        this.first=first;
        this.second=second;
    }
}
class GFG{
 
// Function to print the K most frequent
// elements after each removal
static void maxFreqElements(int arr[], int N, int K)
{
    // Stores frequency of array elements
    Map<Integer, Integer> mp=new HashMap<>();
 
    // Pairs array element with frequency
    ArrayList<pair> v=new ArrayList<>();
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
 
        // Count the frequencies
        mp.put(arr[i],mp.getOrDefault(arr[i],0)+1);
 
        // Insert the element with its
        // current frequency into the vector
        v.add(new pair( arr[i], mp.get(arr[i] )));
    }
 
    // Sort the vector according to
    // higher frequency and smaller
    // element if frequency is same
   Collections.sort(v,(a,b)->(a.second != b.second) ? 
                    b.second-a.second:a.first-b.first);
 
    // Print the first K elements
    // of the array
    for (int i = 0; i < K; i++)
        System.out.print(v.get(i).first + " ");
}
  
   // Driver function
    public static void main (String[] args)
    {
     
    // Given array
    int arr[] = { 1, 3, 2, 1, 4, 1 };
 
    // Given K
    int K = 2;
 
    // Size of the array
    int N = arr.length;
 
    maxFreqElements(arr, N, K);
 
    }
}

// This code is contributed by offbeat
Python3 
# Python 3 program for the above approach

# Function to sort the vector
# of vector of pair

# Function to print the K most frequent
# elements after each removal
def maxFreqElements(arr, N, K):
  
    # Stores frequency of array elements
    mp = {}

    # Pairs array element with frequency
    v = []

    # Traverse the array
    for i in range(N):
        # Count the frequencies
        mp[arr[i]] = mp.get(arr[i],0)+1

        # Insert the element with its
        # current frequency into the vector
        v.append([arr[i], mp.get(arr[i],0)])

    # Sort the vector according to
    # higher frequency and smaller
    c = [1, 1]
    
    # element if frequency is same
    v.sort(reverse = True)

    # Print the first K elements
    # of the array   
    for i in range(K):
        print(c[i], end = " ")

# Driver Code
if __name__ == '__main__':
    
    # Given array
    arr = [1, 3, 2, 1, 4, 1]

    # Given K
    K = 2

    # Size of the array
    N = len(arr)

    maxFreqElements(arr, N, K)
    
    # This code is contributed by SURANDRA_GANGWAR.
C# 
// C# program for above approach
using System;
using System.Collections.Generic;

public class pair
{
  public int first, second;

  public pair(int first, int second)
  {
    this.first = first;
    this.second = second;
  }
}

public class GFG{

  static int Compare(KeyValuePair<int, int> a, KeyValuePair<int, int> b)
  {
    if(a.Value != b.Value)
    {
      return b.Value - a.Value;
    }
    else
    {
      return a.Key - b.Key;
    }
  }

  // Function to print the K most frequent
  // elements after each removal
  static void maxFreqElements(int[] arr, int N, int K)
  {
    
    // Stores frequency of array elements
    Dictionary<int,int> mp = new Dictionary<int,int>();

    // Pairs array element with frequency
    List<KeyValuePair<int,int>> v = new List<KeyValuePair<int,int>>();
    
    // Traverse the array
    for (int i = 0; i < N; i++)
    {

      // Count the frequencies
      if(!mp.ContainsKey(arr[i]))
      {
        mp.Add(arr[i], 1);
      }
      else
      {
        mp[arr[i]]++;
      }

      // Insert the element with its
      // current frequency into the vector
      v.Add(new KeyValuePair<int,int>( arr[i], mp[arr[i] ]));
    }
    v.Sort(Compare);

    // Print the first K elements
    // of the array
    for (int i = 0; i < K; i++)
    {
      Console.Write(v[i].Key + " ");
    }
  }

  // Driver function
  static public void Main ()
  {

    // Given array
    int[] arr = { 1, 3, 2, 1, 4, 1 };

    // Given K
    int K = 2;

    // Size of the array
    int N = arr.Length;

    maxFreqElements(arr, N, K);
  }
}

// This code is contributed by rag2127.
JavaScript 
<script>


// Javascript program for the above approach

// Function to print the K most frequent
// elements after each removal
function maxFreqElements(arr, N, K)
{
    // Stores frequency of array elements
    var mp = new Map();

    // Pairs array element with frequency
    var v = [];

    // Traverse the array
    for (var i = 0; i < N; i++) {

        // Count the frequencies
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1)
        else
            mp.set(arr[i], 1)

        // Insert the element with its
        // current frequency into the vector
        v.push([arr[i], mp.get(arr[i])]);
    }

    // Sort the vector according to
    // higher frequency and smaller
    // element if frequency is same
    v.sort();

    // Print the first K elements
    // of the array
    for (var i = 0; i < K; i++)
        document.write( v[i][0] + " ");
}

// Driver Code
// Given array
var arr = [1, 3, 2, 1, 4, 1];
// Given K
var K = 2;
// Size of the array
var N = arr.length
maxFreqElements(arr, N, K);


</script>

Output: 
1 1

 

Time Complexity: O(NlogN)
Auxiliary Space: O(N)


 


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