Given a string of lowercase letters, reduce it by removing the characters which appear more than k times in the string.
Examples:
Input: str = "neveropen"
k = 2
Output: for
Input: str = "neveropen"
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[] = "neveropen"; 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 timesimport 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[] = "neveropen".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 timesMAX_CHAR = 26def 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 codeif __name__ == '__main__': str = "neveropen" 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 = "neveropen".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$str = "neveropen";$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 = "neveropen"; 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 timesstring 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 codeint main(){ string str = "neveropen"; 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 = "neveropen"; 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 timesfrom collections import Counter# Function to reduce the string by# removing the characters which# appears more than k timesdef 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 Codeif __name__ == "__main__": str = "neveropen" 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 = "neveropen"; int k = 2; Console.WriteLine(RemoveChars(str, k)); }} |
Javascript
// Function to reduce the string by// removing the characters which// appears more than k timesfunction 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 Codelet str = "neveropen";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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