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++

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// 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;
}

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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

[tabbyending]
Output:


1 3 3 4



Time Complexity – O(N)



          
          
          
          

            

    

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