Given a string of lowercase letters and a number K. The task is to reduce it by removing the characters which appear strictly less than K times in the string.
Examples:
Input : str = "neveropen", K = 2 Output : neveropenneveropen Input : str = "neveropen", 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 <bits/stdc++.h>using namespace std;const int MAX_CHAR = 26;// Function to reduce the string by// removing the characters which// appears less than k timesstring 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 Codeint main(){ string str = "neveropen"; 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 timesclass 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 = "neveropen"; 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 timesMAX_CHAR = 26# Function to reduce the string by# removing the characters which# appears less than k timesdef 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 Codeif __name__ == "__main__": str = "neveropen" 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 timesusing 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 = "neveropen"; int k = 2; Console.WriteLine(removeChars(str, k)); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to reduce the string by// removing the characters which// appears less than k timeslet MAX_CHAR = 26;// Function to reduce the string by// removing the characters which// appears less than k timesfunction removeChars(str,k){ // Hash table initialised to 0 let hash = new Array(MAX_CHAR); for(let i=0;i<MAX_CHAR;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 res = ""; 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) { res += str[i]; } } return res;}// Driver Codelet str = "neveropen";let k = 2;document.write(removeChars(str, k));// This code is contributed by rag2127</script> |
Output
neveropenneveropen
Time Complexity: O(N), where N is the length of the given string.
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 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 appear less than k times.
Note: This method is applicable to all types of characters.
Below is the implementation of the above approach:
C++
#include <iostream>#include <unordered_map>using namespace std;string removeChars(string str, int k) { // Using an 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++) { // Append the characters which appear // more than or equal to k times if (freq[str[i]] >= k) { res += str[i]; } } return res;}int main() { string str = "neveropen"; int k = 2; cout << removeChars(str, k) << endl; return 0;} |
Java
import java.util.HashMap;class Main { // Function to reduce the string by // removing the characters which // appear less than k times public static String removeChars(String str, int k) { // Using a HashMap to count frequencies HashMap<Character, Integer> freq = new HashMap<Character, Integer>(); 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++) { // Append the characters which // appear more than or equal to k times if (freq.get(str.charAt(i)) >= k) { res.append(str.charAt(i)); } } 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 less than k timesfrom collections import Counter# Function to reduce the string by# removing the characters which# appears less 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 more 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 Program { static string RemoveChars(string str, int k) { // Using a 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++; } else { freq = 1; } } // Create a new empty string string res = ""; for (int i = 0; i < str.Length; i++) { // Append the characters which appear // more than or equal to k times if (freq[str[i]] >= k) { res += str[i]; } } return res; } 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 less than k timesfunction removeChars(str, k) { // Using an object to count frequencies let freq = {}; for (let i = 0; i < str.length; i++) { if (freq[str[i]]) { freq[str[i]]++; } else { freq[str[i]] = 1; } } // Create a new empty string let res = ""; for (let i = 0; i < str.length; i++) { // Append the characters which // appears more 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 contriibuted by codebraxnzt |
Output
neveropenneveropen
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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