# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ 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 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 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 `

 `# 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# 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 `

Output:
```1 3 3 4
```

Time Complexity – O(N)

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