Remove all occurrences of a character in a string

Given a string. Write a program to remove all the occurrences of a character in the string.

Examples:

Input : s = "geeksforgeeks"
        c = 'e'
Output : s = "gksforgks"

Input : s = "geeksforgeeks"
        c = 'g'
Output : s = "eeksforeeks"

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[] = "geeksforgeeks";
    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 Code
public static void main(String[] args)
{
    String s = "geeksforgeeks";
    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 character
def 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 code
if __name__ == '__main__' :

    s = "geeksforgeeks"
    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 Code
public static void Main()
{
    string s = "geeksforgeeks";
    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
$s = "geeksforgeeks";
removeChar($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 Code
let s = "geeksforgeeks";
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 code
int main()
{
    string word="geeksforgeeks";
    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 = "geeksforgeeks";
        char ch = 'e';
        System.out.println(removechar(word, ch));
    }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python3 code for above approach
def 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 code
word="geeksforgeeks"
ch='e'
print(removechar(word,ch))

# This code is contributed by Akshay Tripathi

C#

// C# code
using 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 = "geeksforgeeks";
        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="geeksforgeeks";
    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).