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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Python3














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# Python3 program to remove the elements which 


# appear strictly less 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 = dict() 


  


    for i in range(n): 


          


        # Incrementing the frequency 


        # of the element by 1. 


        mp[arr[i]] = mp.get(arr[i], 0) + 1


  


    for i in range(n): 


  


        # Print the element which appear 


        # more than or equal to k times. 


        if (arr[i] in mp and mp[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 Mohit Kumar 



















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C#














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// C# program to remove the elements which  


// appear strictly less than k times from the array.  


using System; 


using System.Collections.Generic; 


   


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. 


        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.Add(arr[i], 1); 


            else


            { 


                int x = mp[arr[i]]; 


                mp[arr[i]] = mp[arr[i]] + ++x; 


            } 


        } 


   


        for (int i = 0; i < n; ++i)  


        { 


               


            // Print the element which appear 


            // more than or equal to k times. 


            if (mp[arr[i]] >= k) 


                Console.Write(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 



















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Output:


2 2 3 2 3





Time Complexity – O(N)


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