Remove elements from the array which appear more than k times

Given an array of integers, remove all the occurrences of those elements which appear strictly more than k times in the array.
Examples:

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

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

Approach:

• Take a hash map, which will store the frequency of all the elements in the array.
• Now, traverse once again.
• Print the elements which appear less than or equal to k times.

C++

// C++ program to remove the elements which
// appear more than k times from the array.
#include "iostream"
#include "unordered_map"
using namespace std;

void RemoveElements(int arr[], int n, int k)
{
    // Hash map which will store the
    // frequency of the elements of the array.
    unordered_map<int, int> mp;

    for (int i = 0; i < n; ++i) {
        // Incrementing the frequency
        // of the element by 1.
        mp[arr[i]]++;
    }

    for (int i = 0; i < n; ++i) {
        // Print the element which appear
        // less than or equal to k times.
        if (mp[arr[i]] <= k) {
            cout << arr[i] << " ";
        }
    }
}

int main(int argc, char const* argv[])
{
    int arr[] = { 1, 2, 2, 3, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);

    int k = 2;

    RemoveElements(arr, n, k);
    return 0;
}

Java

// Java program to remove the elements which
// appear more than k times from the array.
import java.util.HashMap;
import java.util.Map;

class GFG
{

static void RemoveElements(int arr[], int n, int k)
{
    // Hash map which will store the
    // frequency of the elements of the array.
    Map<Integer,Integer> mp = new HashMap<>();

    for (int i = 0; i < n; ++i)
    {
        // Incrementing the frequency
        // of the element by 1.
        mp.put(arr[i],mp.get(arr[i]) == null?1:mp.get(arr[i])+1);

    }

    for (int i = 0; i < n; ++i)
    {
        // Print the element which appear
        // less than or equal to k times.
        if (mp.containsKey(arr[i]) && mp.get(arr[i]) <= k)
        {
            System.out.print(arr[i] + " ");
        }
    }
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 2, 3, 2, 3, 4 };
    int n = arr.length;

    int k = 2;

    RemoveElements(arr, n, k);

}
}

// This code is contributed by Rajput-Ji

Python3

# Python 3 program to remove the elements which
# appear more than k times from the array.
def RemoveElements(arr, n, k):

    # Hash map which will store the
    # frequency of the elements of the array.
    mp = {i:0 for i in range(len(arr))}

    for i in range(n):

        # Incrementing the frequency
        # of the element by 1.
        mp[arr[i]] += 1

    for i in range(n):

        # Print the element which appear
        # less than or equal to k times.
        if (mp[arr[i]] <= k):
            print(arr[i], end = " ")

# Driver Code
if __name__ == '__main__':
    arr = [1, 2, 2, 3, 2, 3, 4]
    n = len(arr)

    k = 2

    RemoveElements(arr, n, k)

# This code is contributed by
# Sahil_Shelangia

C#

// C# program to remove the elements which
// appear more than k times from the array.
using System;
using System.Collections.Generic;

class GFG
{
static void RemoveElements(int [] arr,
                           int n, int k)
{
    // Hash map which will store the
    // frequency of the elements of the array.
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();

    for (int i = 0; i < n; ++i)
    {
        // Incrementing the frequency
        // of the element by 1.
        if(mp.ContainsKey(arr[i]))
            mp[arr[i]]++;
        else
            mp[arr[i]] = 1;
    }

    for (int i = 0; i < n; ++i)
    {
        // Print the element which appear
        // less than or equal to k times.
        if (mp.ContainsKey(arr[i]) && mp[arr[i]] <= k)
        {
            Console.Write(arr[i] + " ");
        }
    }
}

// Driver code
static public void Main()
{
    int [] arr = { 1, 2, 2, 3, 2, 3, 4 };
    int n = arr.Length;

    int k = 2;

    RemoveElements(arr, n, k);
}
}

// This code is contributed by Mohit kumar 29

Javascript

<script>

// JavaScript program to remove the elements which
// appear more than k times from the array.

function RemoveElements(arr,n,k)
{
    // Hash map which will store the
    // frequency of the elements of the array.
    let mp = new Map();

    for (let i = 0; i < n; ++i)
    {
        // Incrementing the frequency
        // of the element by 1.
        mp.set(arr[i],mp.get(arr[i]) == null?1:mp.get(arr[i])+1);

    }

    for (let i = 0; i < n; ++i)
    {
        // Print the element which appear
        // less than or equal to k times.
        if (mp.has(arr[i]) && mp.get(arr[i]) <= k)
        {
            document.write(arr[i] + " ");
        }
    }
}

// Driver code
let arr=[1, 2, 2, 3, 2, 3, 4 ];
let n = arr.length;
let k = 2;
RemoveElements(arr, n, k);


// This code is contributed by unknown2108

</script>

Output:

1 3 3 4

Time Complexity – O(N), where N is the size of the given integer.

Auxiliary Space – O(N), where N is the size of the given integer.

Method #2:Using Built-in Python functions:

• Count the frequencies of every element using Counter function
• Traverse the array.
• Print the elements which appear less than or equal to k times.

Below is the implementation of the above approach:

Python3

# Python3 program to remove the elements which
# appear strictly less than k times from the array.
from collections import Counter

def removeElements(arr, n, k):

    # Calculating frequencies
    # using Counter function
    freq = Counter(arr)
    for i in range(n):

        # Print the element which appear
        # more than or equal to k times.
        if (freq[arr[i]] <= k):
            print(arr[i], end=" ")


# Driver Code
arr = [1, 2, 2, 3, 2, 3, 4]
n = len(arr)
k = 2
removeElements(arr, n, k)

# This code is contributed by vikkycirus

Output:

1 3 3 4 

Time Complexity – O(N), where N is the size of the given integer.

Auxiliary Space – O(N), where N is the size of the given integer.