# 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++ 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

2 2 3 2 3

**Time Complexity** – O(N)

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