# 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 = 2Output :1 3 3 4Input :arr[] = {2, 5, 5, 7} k = 1Output :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)

#### 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

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