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 appear strictly less than k times.
• Else, print it.

Implementation:

C++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

<script>

// JavaScript program to remove the elements which
// appear strictly less 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.
            if (!mp.has(arr[i]))
                mp.set(arr[i], 1);
            else
            {
                let x = mp.get(arr[i]);
                mp.set(arr[i], ++x);
            }
        }

        for (let i = 0; i < n; ++i)
        {

            // Print the element which appear
            // more than or equal to k times.
            if (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

2 2 3 2 3 

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

Method #2:Using Built-in Python functions:

• Count the frequencies of every element using Counter() function
• Traverse the array.
• Remove the elements which appear strictly less than k times.
• Else, print it.

Implementation:

C++

// C++ program to remove elements which appear more than k
// times in an array
#include <iostream>
#include <unordered_map>

using namespace std;

void removeElements(int arr[], int n, int k)
{

  // Creating an unordered_map to store the frequencies
  unordered_map<int, int> freq;
  for (int i = 0; i < n; i++) {
    if (freq.count(arr[i])) {
      freq[arr[i]]++;
    }
    else {
      freq[arr[i]] = 1;
    }
  }

  // Printing the elements which appear more than or equal
  // to k times
  for (int i = 0; i < n; i++) {
    if (freq[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;
}

Java

import java.util.*;

public class Main {

  public static void removeElements(int[] arr, int n, int k) {

    // Creating a HashMap to store the frequencies
    HashMap<Integer, Integer> freq = new HashMap<>();
    for (int i = 0; i < n; i++) {
      if (freq.containsKey(arr[i])) {
        freq.put(arr[i], freq.get(arr[i]) + 1);
      }
      else {
        freq.put(arr[i], 1);
      }
    }

    // Printing the elements which appear more than or equal to k times
    for (int i = 0; i < n; i++) {
      if (freq.get(arr[i]) >= k) {
        System.out.print(arr[i] + " ");
      }
    }
  }

  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);
  }
}

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

C#

// C# Equivalent
using System;
using System.Collections.Generic;

public class Program
{
  public static void removeElements(int[] arr, int n, int k)
  {

    // Creating a Dictionary to store the frequencies
    Dictionary<int, int> freq = new Dictionary<int, int>();
    for (int i = 0; i < n; i++)
    {
      if (freq.ContainsKey(arr[i]))
      {
        freq[arr[i]] = freq[arr[i]] + 1;
      }
      else
      {
        freq.Add(arr[i], 1);
      }
    }

    // Printing the elements which appear more than or equal to k times
    for (int i = 0; i < n; i++)
    {
      if (freq[arr[i]] >= k)
      {
        Console.Write(arr[i] + " ");
      }
    }
  }

  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);
  }
}

Javascript

// JavaScript program to remove the elements which
// appear strictly less than k times from the array.
let x="";
function removeElements(arr, n, k) {

  // Calculating frequencies using
  // the reduce() function
  const freq = arr.reduce((count, val) => {
    count[val] = (count[val] || 0) + 1;
    return count;
  }, {});

  for (let i = 0; i < n; i++) {
    // Print the element which appears
    // more than or equal to k times.
    if (freq[arr[i]] >= k) {
      x = x+ arr[i] + " ";
    }
  }
  console.log(x);
}

// Driver Code
const arr = [1, 2, 2, 3, 2, 3, 4];
const n = arr.length;
const k = 2;
removeElements(arr, n, k);

Output

2 2 3 2 3 

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #3:( Space optimization):  First  we will sort the array  and then check if the frequency of array element is greater than or equal to k  with the use of binary search function ( Upper_bound ) . Frequency of array element is ‘last_index-first_index+1’ . If the frequency is greater than one , then print it .      

Below is the implementation of the above approach: 

C++

// C++ implementation of the above approach

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

// Function to remove element that occurs
// lees than k times in the array
void removeElements(int arr[], int n, int k)
{
    sort(arr, arr + n); // sort array for binary search

    for (int i = 0; i < n; i++) {
        // index of first and last occ of arr[i]
        int first_index
            = lower_bound(arr, arr + n, arr[i]) - arr;
        int last_index
            = upper_bound(arr, arr + n, arr[i]) - arr - 1;

        int fre = last_index - first_index
                  + 1; // finding frequency
        if (fre >= k) { // printing element if it is greater
                        // than 1
            cout << arr[i] << " ";
        }
    }
}

// Drive code
int main()
{
    int arr[] = { 1, 2, 2, 3, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;

    // Function call
    removeElements(arr, n, k);
    return 0;
}

// This Code is contributed by nikhilsainiofficial546

Javascript

// javascript implementation of the above approach
function lower_bound(a, x){
    let l = 0;
    let h = arr.length - 1;

    while(l <= h){
        let m = Math.floor((l + h)/2);
        if(a[m] < x){
            l = m + 1;
        }
        else if(a[m] >= x){
            h = m - 1;
        }
    }

    return l;
}

function upper_bound(a, x){
    let l = 0;
    let h = arr.length - 1;

    while(l <= h){
        let m = Math.floor((l + h)/2);

        if(a[m] <= x){
            l = m + 1;
        }
        else if(a[m] > x){
            h = m - 1;
        }
    }

    return l;
}

//Function to remove element that occurs
// lees than k times in the array
function removeElements(arr, n, k)
{
    arr.sort();

    for(let i = 0 ; i < n ;i++)
    {
      //index of first and last occ of arr[i]
      let first_index = lower_bound(arr, arr[i]);
      let last_index = upper_bound(arr, arr[i])- 1;

      let fre = last_index-first_index+1;//finding frequency
      if(fre >= k)
      {
          // printing element if it is greater than 1\
          process.stdout.write(arr[i] + " ");
      }
    }
}

// Drive code
let arr = [1, 2, 2, 3, 2, 3, 4];
let n = arr.length;
let k = 2;

// Function call
removeElements(arr, n, k);

 // The code is contributed by Nidhi goel.

Python3

# Function to remove element that occurs
# less than k times in the array


def removeElements(arr, n, k):
    arr.sort()  # sort array for binary search

    for i in range(n):
        # index of first and last occ of arr[i]
        first_index = bisect.bisect_left(arr, arr[i])
        last_index = bisect.bisect_right(arr, arr[i]) - 1

        fre = last_index - first_index + 1  # finding frequency
        if fre >= k:
            # printing element if it is greater than 1
            print(arr[i], end=" ")


# Driver code
if __name__ == "__main__":
    import bisect

    arr = [1, 2, 2, 3, 2, 3, 4]
    n = len(arr)
    k = 2

    # Function call
    removeElements(arr, n, k)

Java

import java.util.*;

class Main {
    // Function to remove element that occurs
    // lees than k times in the array
    static void removeElements(int arr[], int n, int k)
    {
        Arrays.sort(arr); // sort array for binary search
        ArrayList<Integer> arr1 = new ArrayList<Integer>(5);

        // using add() to initialize values

        for (int i = 0; i < n; i++) {
            arr1.add(arr[i]);
        }
        for (int i = 0; i < n; i++) {
            // index of first and last occ of arr[i]
            int first_index = arr1.indexOf(arr[i]);
            int last_index = arr1.lastIndexOf(arr[i]);
            while (last_index < n
                   && arr[last_index] == arr[i])
                last_index++;
            last_index--;

            int fre = last_index - first_index
                      + 1; // finding frequency
            if (fre >= k) { // printing element if it is
                            // greater than 1
                System.out.print(arr[i] + " ");
            }
        }
    }

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

        // Function call
        removeElements(arr, n, k);
    }
}
// This code is contributed by Pratik Gagare (pratikg03)

C#

using System;
using System.Collections.Generic;
using System.Linq;

class MainClass {
    // Function to remove element that occurs
    // less than k times in the array
    static void RemoveElements(int[] arr, int n, int k)
    {
        Array.Sort(arr); // sort array for binary search
        List<int> arr1 = new List<int>(5);

        // using Add() to initialize values
        for (int i = 0; i < n; i++) {
            arr1.Add(arr[i]);
        }

        for (int i = 0; i < n; i++) {
            // index of first and last occ of arr[i]
            int first_index = arr1.IndexOf(arr[i]);
            int last_index = arr1.LastIndexOf(arr[i]);
            while (last_index < n && arr[last_index] == arr[i])
                last_index++;
            last_index--;

            int fre = last_index - first_index + 1; // finding frequency
            if (fre >= k) { // printing element if it is greater than 1
                Console.Write(arr[i] + " ");
            }
        }
    }

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

        // Function call
        RemoveElements(arr, n, k);
    }
}

Output

2 2 2 3 3 

Time Complexity: O(n*log₂n) 
Auxiliary Space: O(1)