Given a string. Write a program to remove all the occurrences of a character in the string.
Examples:
Input : s = "neveropen"
c = 'e'
Output : s = "gksforgks"
Input : s = "neveropen"
c = 'g'
Output : s = "eeksforeeks"
Input : s = "neveropen"
c = 'k'
Output : s = "geesforgees"
First Approach: The idea is to maintain an index of the resultant string.
Implementation:
C++
// C++ program to remove a particular character// from a string.#include <bits/stdc++.h>using namespace std;void removeChar(char* s, char c){ int j, n = strlen(s); for (int i = j = 0; i < n; i++) if (s[i] != c) s[j++] = s[i]; s[j] = '\0';}int main(){ char s[] = "neveropen"; removeChar(s, 'g'); cout << s; return 0;} |
Java
// Java program to remove // a particular character// from a string.class GFG {static void removeChar(String s, char c){ int j, count = 0, n = s.length(); char []t = s.toCharArray(); for (int i = j = 0; i < n; i++) { if (t[i] != c) t[j++] = t[i]; else count++; } while(count > 0) { t[j++] = '\0'; count--; } System.out.println(t);}// Driver Codepublic static void main(String[] args) { String s = "neveropen"; removeChar(s, 'g');}}// This code is contributed// by ChitraNayal |
Python3
# Python3 program to remove# a particular character# from a string.# function for removing the # occurrence of characterdef removeChar(s, c) : # find total no. of # occurrence of character counts = s.count(c) # convert into list # of characters s = list(s) # keep looping until # counts become 0 while counts : # remove character # from the list s.remove(c) # decremented by one counts -= 1 # join all remaining characters # of the list with empty string s = '' . join(s) print(s)# Driver codeif __name__ == '__main__' : s = "neveropen" removeChar(s,'g') # This code is contributed # by Ankit Rai |
C#
// C# program to remove a// particular character// from a string.using System;class GFG {static void removeChar(string s, char c){ int j, count = 0, n = s.Length; char[] t = s.ToCharArray(); for (int i = j = 0; i < n; i++) { if (s[i] != c) t[j++] = s[i]; else count++; } while(count > 0) { t[j++] = '\0'; count--; } Console.Write(t);}// Driver Codepublic static void Main(){ string s = "neveropen"; removeChar(s, 'g');}}// This code is contributed// by ChitraNayal |
PHP
<?php // PHP program to remove a// particular character// from a string.function removeChar($s, $c){ $n = strlen($s); $count = 0; for ($i = $j = 0; $i < $n; $i++) { if ($s[$i] != $c) $s[$j++] = $s[$i]; else $count++; } while($count--) { $s[$j++] = NULL; } echo $s;}// Driver code$s = "neveropen";removeChar($s, 'g');// This code is contributed// by ChitraNayal?> |
Javascript
<script>// Javascript program to remove// a particular character// from a string.function removeChar(s, c){ let j, count = 0, n = s.length; let t = s.split(""); for(let i = j = 0; i < n; i++) { if (t[i] != c) t[j++] = t[i]; else count++; } while (count > 0) { t[j++] = '\0'; count--; } document.write(t.join(""));}// Driver Codelet s = "neveropen";removeChar(s, 'g');// This code is contributed by avanitrachhadiya2155</script> |
Output
eeksforeeks
Complexity Analysis:
- Time Complexity : O(n) where n is length of input string.
- Auxiliary Space : O(1)
Second Approach in C++: We can also use the STL string class and erase function to delete any character at any position using the base addressing (string.begin()).
Implementation:
C++14
#include <bits/stdc++.h>using namespace std;string removechar(string& word,char& ch){ for(int i=0;i<word.length();i++) { if(word[i]==ch){ word.erase(word.begin()+i); i--; } } return word;}// driver's codeint main(){ string word="neveropen"; char ch='e'; cout<<removechar(word,ch); return 0;} |
Java
// Java code import java.util.*;public class GFG { public static String removechar(String word, char ch) { StringBuilder s = new StringBuilder(word); for (int i = 0; i < s.length(); i++) { if (s.charAt(i) == ch) { s.deleteCharAt(i); i--; } } return s.toString(); } // driver's code public static void main(String args[]) { String word = "neveropen"; char ch = 'e'; System.out.println(removechar(word, ch)); }}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python3 code for above approachdef removechar(word, ch): i = 0 while(i < len(word)): if(word[i] == ch): word = word[:i] + word[i+1:] i -= 1 i += 1 return word# driver's codeword="neveropen"ch='e'print(removechar(word,ch))# This code is contributed by Akshay Tripathi |
C#
// C# codeusing System;using System.Text;using System.Collections;class GFG { public static string removechar(string word, char ch) { StringBuilder s = new StringBuilder(word); for (int i = 0; i < s.Length; i++) { if (s[i] == ch) { s.Remove(i, 1); i--; } } return s.ToString(); } // driver's code public static void Main() { string word = "neveropen"; char ch = 'e'; Console.WriteLine(removechar(word, ch)); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
function removechar(word, ch){ let ans=""; for(let i=0;i<word.length;i++) { if(word[i]!=ch){ ans+=word[i]; } } return ans;}// driver's code let word="neveropen"; let ch='e'; document.write(removechar(word,ch)); |
Output
gksforgks
Complexity Analysis:
- Time Complexity: O(n), where n is length of input string.
- Auxiliary Space: O(1).
Third approach : Using built-in replace() function in Python and Javascript.
Implementation:
C++
#include <iostream>#include <string>using namespace std;// function for removing the occurrence of character using replace() functionvoid removeChar(string& word, char ch) { size_t found = word.find(ch); while (found != string::npos) { // loop until no more occurrences of ch found word.erase(found, 1); // remove character at position found found = word.find(ch, found); // search for the next occurrence of ch } cout << word << endl;}// Driver codeint main() { string word = "neveropen"; removeChar(word, 'k'); return 0;} |
Java
public class Main { // function for removing the occurrence of character using replace() function public static void removeChar(String word, char ch) { word = word.replace(Character.toString(ch), ""); System.out.println(word); } // Driver code public static void main(String[] args) { String word = "neveropen"; removeChar(word, 'k'); }}// code by ksam24000 |
Python3
# Python3 program to remove a particular character# function for removing the # occurrence of character using replace() functiondef removeChar(word, ch) : word = word.replace(ch,'') print(word)# Driver codeif __name__ == '__main__' : word = "neveropen" removeChar(word,'k')# This Code is contributed by Pratik Gupta |
C#
using System;class Program { // function for removing the // occurrence of character using replace() function static void removeChar(string word, char ch) { word = word.Replace(ch.ToString(), ""); Console.WriteLine(word); } // Driver code static void Main(string[] args) { string word = "neveropen"; removeChar(word, 'k'); }} |
Javascript
// Javascript program to remove a particular charactervar word = "neveropen";// '/g/' to find the character globally and replace them with empty parametervar c =/k/g ;word = word.replace(c, '');console.log(word); |
Output
geesforgees
Complexity Analysis:
- Time Complexity: O(n), where n is the length of the input string.
- Auxiliary Space: O(1).
Fourth Approach: Using Recursion
Here our recursive function’s base case will be when string length’s become 0. And in the recursive function, if the encountered character is the character that we have to remove then call the recursive function from the next character else store this character in answer and then call the recursive function from the next character.
Code:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to remove all occurrences// of a character in the stringstring removeChar(string str, char ch){ // Base Case if (str.length() == 0) { return ""; } // Check the first character // of the given string if (str[0] == ch) { // Pass the rest of the string // to recursion Function call return removeChar(str.substr(1), ch); } // Add the first character of str // and string from recursion return str[0] + removeChar(str.substr(1), ch);}// Driver Codeint main(){ // Given String string str = "neveropen"; // Function Call str = removeChar(str, 'g'); cout << str; return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class Main { // Function to remove all occurrences // of a character in the string static String removeChar(String str, char ch) { // Base Case if (str.length() == 0) { return ""; } // Check the first character // of the given string if (str.charAt(0) == ch) { // Pass the rest of the string // to recursion Function call return removeChar(str.substring(1), ch); } // Add the first character of str // and string from recursion return str.charAt(0) + removeChar(str.substring(1), ch); } // Driver Code public static void main(String[] args) { // Given String String str = "neveropen"; // Function Call str = removeChar(str, 'g'); System.out.println(str); }} |
Python3
# Function to remove all occurrences# of a character in the stringdef removeChar(str, ch): # Base Case if len(str) == 0: return "" # Check the first character # of the given string if str[0] == ch: # Pass the rest of the string # to recursion Function call return removeChar(str[1:], ch) # Add the first character of str # and string from recursion return str[0] + removeChar(str[1:], ch)# Driver Codeif __name__ == '__main__': # Given String str = "neveropen" # Function Call str = removeChar(str, 'g') print(str) |
C#
using System;class MainClass { // Function to remove all occurrences // of a character in the string public static string RemoveChar(string str, char ch) { // Base Case if (str.Length == 0) { return ""; } // Check the first character // of the given string if (str[0] == ch) { // Pass the rest of the string // to recursion Function call return RemoveChar(str.Substring(1), ch); } // Add the first character of str // and string from recursion return str[0] + RemoveChar(str.Substring(1), ch); } // Driver Code public static void Main(string[] args) { // Given String string str = "neveropen"; // Function Call str = RemoveChar(str, 'g'); Console.WriteLine(str); }} |
Javascript
function removeChar(str, ch) { // Base Case if (str.length == 0) { return ""; } // Check the first character // of the given string if (str[0] == ch) { // Pass the rest of the string // to recursion Function call return removeChar(str.slice(1), ch); } // Add the first character of str // and string from recursion return str[0] + removeChar(str.slice(1), ch);}// Driver Codelet str = "neveropen";str = removeChar(str, 'g');console.log(str); |
Output
eeksforeeks
Complexity Analysis-
- Time Complexity: O(n), where n is the length of the input string.
- Auxiliary Space: O(n),if we consider recursion stack space
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