Remove elements from the array whose frequency lies in the range [l, r]
Given an array of integers, remove the elements from the array whose frequency lies in the range [l, r].
Examples:
Input : arr[] = { 1, 2, 3, 3, 2, 2, 5 }
l = 2, r = 3
Output : 1 5
We remove all those elements with frequencies from 2 to 5.
Input : arr[] = { 1, 2, 3, 3, 3, 3 }
l = 2, r = 3
Output : 3 3 3 3
Approach :
- Take a hash map, which will store the frequency of all the elements in the array.
- Now, traverse once again.
- Remove the elements whose frequency lies between [l, r].
- Else, print it.
Implementation:
C++
#include "iostream"
#include "unordered_map"
using namespace std;
void removeElements( int arr[], int n, int l, int r)
{
unordered_map< int , int > mp;
for ( int i = 0; i < n; ++i) {
mp[arr[i]]++;
}
for ( int i = 0; i < n; ++i) {
if (mp[arr[i]] < l or mp[arr[i] > r]) {
cout << arr[i] << " " ;
}
}
}
int main()
{
int arr[] = { 1, 2, 3, 3, 2, 2, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
int l = 2, r = 3;
removeElements(arr, n, l, r);
return 0;
}
|
Java
import java.util.HashMap;
class GFG {
static void removeElements( int arr[], int n, int l, int r)
{
HashMap<Integer, Integer> mp = new HashMap<>();
for ( int i = 0 ; i < n; ++i) {
int val = 0 ;
if (mp.get(arr[i]) == null ) {
val = 1 ;
}
else {
val = mp.get(arr[i]) + 1 ;
}
mp.put(arr[i], val);
}
for ( int i = 0 ; i < n; ++i) {
if (mp.get(arr[i]) < l || mp.get(arr[i]) > r) {
System.out.print(arr[i] + " " );
}
}
}
public static void main(String args[])
{
int arr[] = { 1 , 2 , 3 , 3 , 2 , 2 , 5 };
int n = arr.length;
int l = 2 , r = 3 ;
removeElements(arr, n, l, r);
}
}
|
Python3
def removeElements(arr, n, l, r):
mp = {i: 0 for i in range ( len (arr))}
for i in range (n):
mp[arr[i]] + = 1
for i in range (n):
if (mp[arr[i]] < l or mp[arr[i] > r]):
print (arr[i], end = " " )
if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 , 3 , 2 , 2 , 5 ]
n = len (arr)
l = 2
r = 3
removeElements(arr, n, l, r);
|
C#
using System;
using System.Collections.Generic;
class GFG {
static void removeElements( int [] arr, int n, int l, int r)
{
Dictionary< int , int > mp = new Dictionary< int , int >();
for ( int i = 0; i < n; ++i) {
int val = 0;
if (!mp.ContainsKey(arr[i])) {
val = 1;
}
else {
val = mp[arr[i]] + 1;
}
if (!mp.ContainsKey(arr[i]))
mp.Add(arr[i], val);
else {
mp.Remove(arr[i]);
mp.Add(arr[i], val);
}
}
for ( int i = 0; i < n; ++i) {
if (mp[arr[i]] < l || mp[arr[i]] > r) {
Console.Write(arr[i] + " " );
}
}
}
public static void Main(String[] args)
{
int [] arr = { 1, 2, 3, 3, 2, 2, 5 };
int n = arr.Length;
int l = 2, r = 3;
removeElements(arr, n, l, r);
}
}
|
Javascript
<script>
function removeElements(arr, n, l, r)
{
let mp = new Map();
for (let i = 0; i < n; ++i)
{
let val = 0;
if (mp.get(arr[i]) == null )
{
val = 1;
}
else
{
val = mp.get(arr[i]) + 1;
}
mp.set(arr[i], val);
}
for (let i = 0; i < n; ++i)
{
if (mp.get(arr[i]) < l || mp.get(arr[i]) > r)
{
document.write(arr[i] + " " );
}
}
}
let arr = [ 1, 2, 3, 3, 2, 2, 5 ];
let n = arr.length;
let l = 2, r = 3;
removeElements(arr, n, l, r);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Last Updated :
31 Dec, 2022
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