Remove elements from the array which appear more than k times
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 7
Approach:
- 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
// 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 |
Python3
# 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#
// 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 |
Javascript
<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:
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 |
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.
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