Remove characters that appear more than k times

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Given a string of lowercase letters, reduce it by removing the characters which appear more than k times in the string.

Examples:

Input: str = "zambiatek"
        k = 2
Output: for

Input: str = "zambiatek"
        k = 3
Output: gksforgks

Approach :

Create a hash table of 26 indexes, where the 0th index represents ‘a’ and 1th index represents ‘b’ and so on. Initialize the hash table to zero.
Iterate through the string and count increment the frequency of the str[i] character in the hash table.
Now once again traverse through the string and append those characters in the new string whose frequency in the hash table is less than k and skip those which appear more than equal to k.

Time Complexity: O(N)

Below is the implementation of above approach:

C++

// C++ program to reduce the string by
// removing the characters which
// appears more than k times
#include <bits/stdc++.h>
using namespace std;
 
const int MAX_CHAR = 26;
 
void removeChars(char arr[], int k)
{
    // Hash table initialised to 0
    int hash[MAX_CHAR] = { 0 };
 
    // Increment the frequency of the character
    int n = strlen(arr);
    for (int i = 0; i < n; ++i) 
        hash[arr[i] - 'a']++;    
 
    // Next index in reduced string
    int index = 0;
    for (int i = 0; i < n; ++i) {
 
        // Append the characters which
        // appears less than k times
        if (hash[arr[i] - 'a'] < k) {
           arr[index++] = arr[i];
        }
    }
 
    arr[index] = '\0';
}
 
int main()
{
    char str[] = "zambiatek";
    int k = 2;
    removeChars(str, k);
    cout << str;
    return 0;
}

Java

// Java program to reduce the string by
// removing the characters which
// appears more than k times
import java.util.*;
class Solution
{
  
static final int MAX_CHAR = 26;
  
static void removeChars(char arr[], int k)
{
    // Hash table initialised to 0
    int hash[]=new int[MAX_CHAR];
     
    for (int i = 0; i <MAX_CHAR; ++i)
    hash[i]=0;
  
    // Increment the frequency of the character
    int n = (arr).length;
    for (int i = 0; i < n; ++i) 
        hash[arr[i] - 'a']++;    
  
    // Next index in reduced string
    int index = 0;
    for (int i = 0; i < n; ++i) {
  
        // Append the characters which
        // appears less than k times
        if (hash[arr[i] - 'a'] < k) {
           arr[index++] = arr[i];
        }
    }
    for (int i = index; i < n; ++i)
    arr[i] = ' ';
}
  
public static void main(String args[])
{
    char str[] = "zambiatek".toCharArray();;
    int k = 2;
    removeChars(str, k);
    System.out.println(String.valueOf( str));
}
 
}
//contributed by Arnab Kundu

Python3

# Python 3 program to reduce the string by
# removing the characters which
# appears more than k times
MAX_CHAR = 26
 
def removeChars(arr, k):
     
    # Hash table initialised to 0
    hash = [0 for i in range(MAX_CHAR)]
 
    # Increment the frequency of the character
    n = len(arr)
    for i in range(n):
        hash[ord(arr[i]) - ord('a')] += 1
 
    # Next index in reduced string
    index = 0
    for i in range(n):
         
        # Append the characters which
        # appears less than k times
        if (hash[ord(arr[i]) - ord('a')] < k):
            arr[index] = arr[i]
            index += 1
         
    arr[index] = ' '
     
    for i in range(index):
        print(arr[i], end = '')
 
# Driver code
if __name__ == '__main__':
    str = "zambiatek"
    str = list(str)
    k = 2
    removeChars(str, k)
 
# This code is contributed by
# Shashank_Sharma

C#

// C# program to reduce the string by 
// removing the characters which 
// appears more than k times 
using System;
public class Solution{ 
 
    static readonly int MAX_CHAR = 26; 
 
    static void removeChars(char []arr, int k) 
    { 
        // Hash table initialised to 0 
        int []hash=new int[MAX_CHAR]; 
 
        for (int i = 0; i <MAX_CHAR; ++i) 
        hash[i]=0; 
 
        // Increment the frequency of the character 
        int n = (arr).Length; 
        for (int i = 0; i < n; ++i) 
            hash[arr[i] - 'a']++;     
 
        // Next index in reduced string 
        int index = 0; 
        for (int i = 0; i < n; ++i) { 
 
            // Append the characters which 
            // appears less than k times 
            if (hash[arr[i] - 'a'] < k) { 
            arr[index++] = arr[i]; 
            } 
        } 
        for (int i = index; i < n; ++i) 
        arr[i] = ' '; 
    } 
 
    public static void Main() 
    { 
        char []str = "zambiatek".ToCharArray();; 
        int k = 2; 
        removeChars(str, k); 
        Console.Write(String.Join("",str)); 
    } 
 
} 
// This code is contributed by PrinciRaj1992

PHP

<?php 
// PHP program to reduce the string by
// removing the characters which
// appears more than k times
$MAX_CHAR = 26;
 
function removeChars($arr, $k)
{
    global $MAX_CHAR;
     
    // Hash table initialised to 0
    $hash = array_fill(0, $MAX_CHAR, NULL);
 
    // Increment the frequency of 
    // the character
    $n = strlen($arr);
    for ($i = 0; $i < $n; ++$i) 
        $hash[ord($arr[$i]) - ord('a')]++; 
 
    // Next index in reduced string
    $index = 0;
    for ($i = 0; $i < $n; ++$i)
    {
 
        // Append the characters which
        // appears less than k times
        if ($hash[ord($arr[$i]) - ord('a')] < $k)
        {
            $arr[$index++] = $arr[$i];
        }
    }
     
    $arr[$index] = '';
    for($i = 0; $i < $index; $i++)
        echo $arr[$i];
}
 
// Driver Code
$str = "zambiatek";
$k = 2;
removeChars($str, $k);
 
// This code is contributed by ita_c
?>

Javascript

<script>
 
// JavaScript program to reduce the string by
// removing the characters which
// appears less than k times
     
    let  MAX_CHAR = 26;
     
    // Function to reduce the string by
// removing the characters which
// appears less than k times
    function removeChars(str,k)
    {
        // Hash table initialised to 0
        let hash = new Array(MAX_CHAR);
         for(let i=0;i<hash.length;i++)
        {
            hash[i]=0;
        }
        // Increment the frequency of the character
        let n = str.length;
        for (let i = 0; i < n; ++i) {
            hash[str[i].charCodeAt(0) - 
            'a'.charCodeAt(0)]++;
        }
  
        // create a new empty string
        let index = "";
        for (let i = 0; i < n; ++i) {
  
            // Append the characters which
            // appears more than equal to k times
            if (hash[str[i].charCodeAt(0) - 
            'a'.charCodeAt(0)] < k) 
            {
                index += str[i];
            }
        }
  
        return index;
    }
     
    // Driver Code
    let str = "zambiatek";
    let k = 2;
    document.write(removeChars(str, k));
     
 
// This code is contributed by rag2127
 
</script>

Output

for

Complexity Analysis:

Time Complexity: O(n), where n represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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 once again traverse through the string and append those characters in the new string whose frequency in the frequency dictionary is less than k and skip those which appear more than equal to k.

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

Below is the implementation of the above approach:

C++

#include <iostream>
#include <unordered_map>
 
using namespace std;
 
// Function to reduce the string by
// removing the characters which
// appears more than k times
string removeChars(string str, int k)
{
    // Using unordered_map to count frequencies
    unordered_map<char, int> freq;
    for (int i = 0; i < str.length(); i++) {
        char c = str[i];
        freq++;
    }
 
    // create a new empty string
    string res = "";
 
    for (int i = 0; i < str.length(); i++) {
        char c = str[i];
        // Append the characters which
        // appears less than equal to k times
        if (freq < k) {
            res += c;
        }
    }
    return res;
}
 
// Driver code
int main()
{
    string str = "zambiatek";
    int k = 2;
    cout << removeChars(str, k) << endl;
    return 0;
}

Java

import java.util.*;
 
public class Main {
 
    // Function to reduce the string by
    // removing the characters which
    // appears more than k times
    public static String removeChars(String str, int k)
    {
        // Using HashMap to count frequencies
        Map<Character, Integer> freq = new HashMap<>();
        for (int i = 0; i < str.length(); i++) {
            char c = str.charAt(i);
            freq.put(c, freq.getOrDefault(c, 0) + 1);
        }
 
        // create a new empty string
        StringBuilder res = new StringBuilder();
 
        for (int i = 0; i < str.length(); i++) {
            char c = str.charAt(i);
            // Append the characters which
            // appears less than equal to k times
            if (freq.get(c) < k) {
                res.append(c);
            }
        }
        return res.toString();
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "zambiatek";
        int k = 2;
        System.out.println(removeChars(str, k));
    }
}

Python3

# Python 3 program to reduce the string
# by removing the characters which
# appears more than k times
from collections import Counter
 
# Function to reduce the string by
# removing the characters which
# appears more 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 less than equal to k times
        if (freq[str[i]] < k):
            res += str[i]
 
    return res
 
 
# Driver Code
if __name__ == "__main__":
 
    str = "zambiatek"
    k = 2
 
    print(removeChars(str, k))
 
# This code is contributed by vikkycirus

C#

using System;
using System.Collections.Generic;
 
class Gfg
{
    // Function to reduce the string by
    // removing the characters which
    // appears more than k times
    static string RemoveChars(string str, int k)
    {
        // Using Dictionary to count frequencies
        Dictionary<char, int> freq = new Dictionary<char, int>();
        for (int i = 0; i < str.Length; i++)
        {
            char c = str[i];
            if (!freq.ContainsKey(c))
            {
                freq.Add(c, 0);
            }
            freq++;
        }
 
        // create a new empty string
        string res = "";
 
        for (int i = 0; i < str.Length; i++)
        {
            char c = str[i];
            // Append the characters which
            // appears less than equal to k times
            if (freq < k)
            {
                res += c;
            }
        }
        return res;
    }
 
    // Driver code
    static void Main(string[] args)
    {
        string str = "zambiatek";
        int k = 2;
        Console.WriteLine(RemoveChars(str, k));
    }
}

Javascript

// Function to reduce the string by
// removing the characters which
// appears more than k times
function removeChars(str, k)
{
 
    // Using object to count frequencies
    let freq = {};
    for (let i = 0; i < str.length; i++) {
        freq[str[i]] = (freq[str[i]] || 0) + 1;
    }
     
    // create a new empty string
    let res = "";
 
    for (let i = 0; i < str.length; i++)
    {
     
        // Append the characters which
        // appears less than equal to k times
        if (freq[str[i]] < k) {
            res += str[i];
        }
    }
 
    return res;
}
 
// Driver Code
let str = "zambiatek";
let k = 2;
 
console.log(removeChars(str, k));
 
// This code is contributed by codebraxnzt

Output

for

Complexity Analysis:

Time Complexity: O(n), where n represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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