Remove elements that appear strictly less than k times

Given an array of integers, remove all the elements which appear strictly less than k times.

Examples:

Input : arr[] = {1, 2, 2, 3, 2, 3, 4}
        k = 2
Output : 2 2 3 2 3
Explanation : {1, 4} appears less than 2 times.

Approach :

  • Take a hash map, which will store the frequency of all the elements in the array.
  • Now, traverse once again.
  • Remove the elements which appears strictly less than k times.
  • Else, print it.

C++

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// C++ program to remove the elements which
// appear strictly less 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
        // more than or equal to k times.
        if (mp[arr[i]] >= k) {
            cout << arr[i] << " ";
        }
    }
}
  
int main()
{
    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;
}

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Java

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// Java program to remove the elements which 
// appear strictly less than k times from the array. 
import java.util.HashMap;
  
class geeks 
{
  
    public static void removeElements(int[] arr, 
                                        int n, int k)
    {
          
        // Hash map which will store the
        // frequency of the elements of the array.
        HashMap<Integer, Integer> mp = new HashMap<>();
  
        for (int i = 0; i < n; ++i) 
        {
  
            // Incrementing the frequency
            // of the element by 1.
            if (!mp.containsKey(arr[i]))
                mp.put(arr[i], 1);
            else
            {
                int x = mp.get(arr[i]);
                mp.put(arr[i], ++x);
            }
        }
  
        for (int i = 0; i < n; ++i) 
        {
              
            // Print the element which appear
            // more than or equal to k times.
            if (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
// sanjeev2552

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Output:

2 2 3 2 3

Time Complexity – O(N)



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Improved By : sanjeev2552