Remove elements from the array which appear more than k times
Last Updated :
18 Jan, 2024
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 7Approach:
- 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 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
# 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# 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
<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:
C++
#include <iostream>
#include <unordered_map>
#include <vector>
void removeElements(const std::vector<int>& arr, int k) {
// Calculating frequencies using unordered_map
std::unordered_map<int, int> freq;
// Counting the frequency of each element
for (int i = 0; i < arr.size(); ++i) {
freq[arr[i]]++;
}
// Print the elements which appear more than or equal to k times
for (int i = 0; i < arr.size(); ++i) {
if (freq[arr[i]] <= k) {
std::cout << arr[i] << " ";
}
}
}
int main() {
std::vector<int> arr = {1, 2, 2, 3, 2, 3, 4};
int k = 2;
removeElements(arr, k);
return 0;
}
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
public class Main {
public static void
removeElements(ArrayList<Integer> arr, int k)
{
// Calculating frequencies using HashMap
Map<Integer, Integer> freq = new HashMap<>();
// Counting the frequency of each element
for (int i = 0; i < arr.size(); ++i) {
freq.put(arr.get(i),
freq.getOrDefault(arr.get(i), 0) + 1);
}
// Print the elements which appear more than or
// equal to k times
for (int i = 0; i < arr.size(); ++i) {
if (freq.get(arr.get(i)) <= k) {
System.out.print(arr.get(i) + " ");
}
}
}
public static void main(String[] args)
{
ArrayList<Integer> arr = new ArrayList<>();
arr.add(1);
arr.add(2);
arr.add(2);
arr.add(3);
arr.add(2);
arr.add(3);
arr.add(4);
int k = 2;
removeElements(arr, k);
}
}
# 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
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static void RemoveElements(List<int> arr, int k)
{
// Calculating frequencies using Dictionary
Dictionary<int, int> freq = new Dictionary<int, int>();
// Counting the frequency of each element
foreach (int element in arr)
{
if (freq.ContainsKey(element))
{
freq[element]++;
}
else
{
freq[element] = 1;
}
}
// Print the elements which appear more than or equal to k times
foreach (int element in arr)
{
if (freq[element] <= k)
{
Console.Write(element + " ");
}
}
}
static void Main()
{
List<int> arr = new List<int> { 1, 2, 2, 3, 2, 3, 4 };
int k = 2;
RemoveElements(arr, k);
Console.ReadLine();
}
}
function removeElements(arr, k) {
// Calculating frequencies using Map
const freq = new Map();
// Counting the frequency of each element
for (let i = 0; i < arr.length; ++i) {
freq.set(arr[i], (freq.get(arr[i]) || 0) + 1);
}
// Print the elements which appear more than or equal to k times
for (let i = 0; i < arr.length; ++i) {
if (freq.get(arr[i]) <= k) {
console.log(arr[i] + " ");
}
}
}
// Driver code
const arr = [1, 2, 2, 3, 2, 3, 4];
const k = 2;
removeElements(arr, k);
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.
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