Related Articles

# Remove characters from string that appears strictly less than K times

• Last Updated : 21 Jul, 2021

Given a string of lowercase letters and a number K. The task is to reduce it by removing the characters which appears strictly less than K times in the string.
Examples

```Input : str = "geeksforgeeks", K = 2
Output : geeksgeeks

Input : str = "geeksforgeeks", K = 3
Output : eeee```

Approach :

• Create a hash table of 26 indexes, where 0th index representing ‘a’ and 1th index represent ‘b’ and so on to store the frequency of each of the characters in the input string. Initialize this hash table to zero.
• Iterate through the string and increment the frequency of each character in the hash table. That is, hash[str[i]-‘a’]++.
• Now create a new empty string and once again traverse through the input string and append-only those characters in the new string whose frequency in the hash table is more than or equal to k and skip those which appear less than k times.

Below is the implementation of the above approach:

## C++

 `// C++ program to reduce the string by``// removing the characters which``// appears less than k times` `#include ``using` `namespace` `std;` `const` `int` `MAX_CHAR = 26;` `// Function to reduce the string by``// removing the characters which``// appears less than k times``string removeChars(string str, ``int` `k)``{``    ``// Hash table initialised to 0``    ``int` `hash[MAX_CHAR] = { 0 };` `    ``// Increment the frequency of the character``    ``int` `n = str.length();``    ``for` `(``int` `i = 0; i < n; ++i)``        ``hash[str[i] - ``'a'``]++;` `    ``// create a new empty string``    ``string res = ``""``;``    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Append the characters which``        ``// appears more than equal to k times``        ``if` `(hash[str[i] - ``'a'``] >= k) {``            ``res += str[i];``        ``}``    ``}` `    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``string str = ``"geeksforgeeks"``;``    ``int` `k = 2;` `    ``cout << removeChars(str, k);` `    ``return` `0;``}`

## Java

 `// Java program to reduce the string by``// removing the characters which``// appears less than k times``class` `GFG {` `    ``final` `static` `int` `MAX_CHAR = ``26``;` `// Function to reduce the string by``// removing the characters which``// appears less than k times``    ``static` `String removeChars(String str, ``int` `k) {``        ``// Hash table initialised to 0``        ``int` `hash[] = ``new` `int``[MAX_CHAR];` `        ``// Increment the frequency of the character``        ``int` `n = str.length();``        ``for` `(``int` `i = ``0``; i < n; ++i) {``            ``hash[str.charAt(i) - ``'a'``]++;``        ``}` `        ``// create a new empty string``        ``String res = ``""``;``        ``for` `(``int` `i = ``0``; i < n; ++i) {` `            ``// Append the characters which``            ``// appears more than equal to k times``            ``if` `(hash[str.charAt(i) - ``'a'``] >= k) {``                ``res += str.charAt(i);``            ``}``        ``}` `        ``return` `res;``    ``}` `// Driver Code``    ``static` `public` `void` `main(String[] args) {``        ``String str = ``"geeksforgeeks"``;``        ``int` `k = ``2``;` `        ``System.out.println(removeChars(str, k));``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python 3

 `# Python 3 program to reduce the string``# by removing the characters which``# appears less than k times``MAX_CHAR ``=` `26` `# Function to reduce the string by``# removing the characters which``# appears less than k times``def` `removeChars(``str``, k):` `    ``# Hash table initialised to 0``    ``hash` `=` `[``0``] ``*` `(MAX_CHAR)` `    ``# Increment the frequency of``    ``# the character``    ``n ``=` `len``(``str``)``    ``for` `i ``in` `range``(n):``        ``hash``[``ord``(``str``[i]) ``-` `ord``(``'a'``)] ``+``=` `1` `    ``# create a new empty string``    ``res ``=` `""``    ``for` `i ``in` `range``(n):` `        ``# Append the characters which``        ``# appears more than equal to k times``        ``if` `(``hash``[``ord``(``str``[i]) ``-` `ord``(``'a'``)] >``=` `k) :``            ``res ``+``=` `str``[i]` `    ``return` `res` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``str` `=` `"geeksforgeeks"``    ``k ``=` `2` `    ``print``(removeChars(``str``, k))` `# This code is contributed by ita_c`

## C#

 `// C# program to reduce the string by``// removing the characters which``// appears less than k times``using` `System;``class` `GFG``{` `    ``readonly` `static` `int` `MAX_CHAR = 26;` `    ``// Function to reduce the string by``    ``// removing the characters which``    ``// appears less than k times``    ``static` `String removeChars(String str, ``int` `k) ``    ``{``        ``// Hash table initialised to 0``        ``int` `[]hash = ``new` `int``[MAX_CHAR];` `        ``// Increment the frequency of the character``        ``int` `n = str.Length;``        ``for` `(``int` `i = 0; i < n; ++i)``        ``{``            ``hash[str[i] - ``'a'``]++;``        ``}` `        ``// create a new empty string``        ``String res = ``""``;``        ``for` `(``int` `i = 0; i < n; ++i)``        ``{` `            ``// Append the characters which``            ``// appears more than equal to k times``            ``if` `(hash[str[i] - ``'a'``] >= k)``            ``{``                ``res += str[i];``            ``}``        ``}` `        ``return` `res;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{``        ``String str = ``"geeksforgeeks"``;``        ``int` `k = 2;` `        ``Console.WriteLine(removeChars(str, k));``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`geeksgeeks`

Time Complexity: O(N), where N is the length of the given string.

#### Method #2:Using Built-in Python functions:

• We will scan the string and count the occurrence of all characters  using built in Counter() function .
• Now create a new empty string and once again traverse through the input string and append only those characters in the new string whose frequency dictionary value is more than or equal to k and skip those which appears less than k times.

Note: This method is applicable for all types of characters.

Below is the implementation of the above approach:

## Python3

 `# Python 3 program to reduce the string``# by removing the characters which``# appears less than k times``from` `collections ``import` `Counter` `# Function to reduce the string by``# removing the characters which``# appears less than k times``def` `removeChars(``str``, k):``  ` `    ``# Using Counter function to``    ``# count frequencies``    ``freq ``=` `Counter(``str``)``    ` `    ``# Create a new empty string``    ``res ``=` `""` `    ``for` `i ``in` `range``(``len``(``str``)):` `        ``# Append the characters which``        ``# appears more than equal to k times``        ``if` `(freq[``str``[i]] >``=` `k):``            ``res ``+``=` `str``[i]` `    ``return` `res`  `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``str` `=` `"geeksforgeeks"``    ``k ``=` `2` `    ``print``(removeChars(``str``, k))` `# This code is contributed by vikkycirus`

Output:

`geeksgeeks`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up