Remove characters from the first string which are present in the second string

Difficulty Level : Easy
Last Updated : 07 Jul, 2021

Write an efficient C function that takes two strings as arguments and removes the characters from the first string which are present in the second string (mask string).

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Algorithm: Let the first input string be a ”test string” and the string which has characters to be removed from the first string be a “mask”

Initialize: res_ind = 0 /* index to keep track of the processing of each character in i/p string */
ip_ind = 0 /* index to keep track of the processing of each character in the resultant string */
Construct count array from mask_str. The count array would be:
(We can use a Boolean array here instead of an int count array because we don’t need a count, we need to know only if the character is present in a mask string)
count[‘a’] = 1
count[‘k’] = 1
count[‘m’] = 1
count[‘s’] = 1
Process each character in the input string and if the count of that character is 0, then only add the character to the resultant string.
str = “tet tringng” // ’s’ has been removed because ’s’ was present in mask_str, but we have got two extra characters “ng”
ip_ind = 11
res_ind = 9
Put a ‘\0′ at the end of the string?

Implementations:

C++

// C++ program to remove duplicates, the order of
// characters is not maintained in this progress
#include <bits/stdc++.h>
#define NO_OF_CHAR 256
using namespace std;
 
int* getcountarray(string str2)
{
    int* count = (int*)calloc(sizeof(int), NO_OF_CHAR);
 
    for (int i = 0; i < str2.size(); i++)
    {
        count[str2[i]]++;
    }
 
    return count;
}
 
/* removeDirtyChars takes two
string as arguments: First
string (str1)  is the one from
where function removes dirty
characters. Second  string(str2)
is the string which contain
all dirty characters which need
to be removed  from first
string */
string removeDirtyChars(string str1, string str2)
{
    // str2 is the string
    // which is to be removed
    int* count = getcountarray(str2);
    string res;
      
    // ip_idx helps to keep
    // track of the first string
    int ip_idx = 0;
 
    while (ip_idx < str1.size())
    {
        char temp = str1[ip_idx];
        if (count[temp] == 0)
        {
            res.push_back(temp);
        }
        ip_idx++;
    }
 
    return res;
}
 
// Driver Code
int main()
{
    string str1 = "geeksforgeeks";
    string str2 = "mask";
 
    // Function call
    cout << removeDirtyChars(str1, str2) << endl;
}

C

#include <stdio.h>
#include <stdlib.h>
#define NO_OF_CHARS 256
 
/* Returns an array of size 256 containing count
   of characters in the passed char array */
int* getCharCountArray(char* str)
{
    int* count = (int*)calloc(sizeof(int), NO_OF_CHARS);
    int i;
    for (i = 0; *(str + i); i++)
        count[*(str + i)]++;
    return count;
}
 
/* removeDirtyChars takes two
string as arguments: First
string (str)  is the one from
where function removes dirty
characters. Second  string is
the string which contain all
dirty characters which need to
be removed  from first string
*/
char* removeDirtyChars(char* str, char* mask_str)
{
    int* count = getCharCountArray(mask_str);
    int ip_ind = 0, res_ind = 0;
    while (*(str + ip_ind))
    {
        char temp = *(str + ip_ind);
        if (count[temp] == 0)
        {
            *(str + res_ind) = *(str + ip_ind);
            res_ind++;
        }
        ip_ind++;
    }
 
    /* After above step string is ngring.
      Removing extra "iittg" after string*/
    *(str + res_ind) = '\0';
 
    return str;
}
 
/* Driver code*/
int main()
{
    char str[] = "geeksforgeeks";
    char mask_str[] = "mask";
    printf("%s", removeDirtyChars(str, mask_str));
    return 0;
}

Java

// Java program to remove duplicates, the order of
// characters is not maintained in this program
 
public class GFG {
    static final int NO_OF_CHARS = 256;
 
    /* Returns an array of size 256 containing count
       of characters in the passed char array */
    static int[] getCharCountArray(String str)
    {
        int count[] = new int[NO_OF_CHARS];
        for (int i = 0; i < str.length(); i++)
            count[str.charAt(i)]++;
 
        return count;
    }
 
    /* removeDirtyChars takes two
    string as arguments: First
    string (str)  is the one from
    where function removes
    dirty characters. Second 
    string is the string which
    contain all dirty characters
    which need to be removed
    from first string */
    static String removeDirtyChars(String str,
                                   String mask_str)
    {
        int count[] = getCharCountArray(mask_str);
        int ip_ind = 0, res_ind = 0;
 
        char arr[] = str.toCharArray();
 
        while (ip_ind != arr.length)
        {
            char temp = arr[ip_ind];
            if (count[temp] == 0) {
                arr[res_ind] = arr[ip_ind];
                res_ind++;
            }
            ip_ind++;
        }
 
        str = new String(arr);
 
        /* After above step string is ngring.
        Removing extra "iittg" after string*/
 
        return str.substring(0, res_ind);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        String mask_str = "mask";
        System.out.println(removeDirtyChars(str, mask_str));
    }
}

Python

# Python program to remove characters
# from first string which
# are present in the second string
NO_OF_CHARS = 256
 
# Utility function to convert
# from string to list
 
 
def toList(string):
    temp = []
    for x in string:
        temp.append(x)
    return temp
 
# Utility function to
# convert from list to string
 
 
def toString(List):
    return ''.join(List)
 
# Returns an array of size
# 256 containing count of characters
# in the passed char array
 
 
def getCharCountArray(string):
    count = [0] * NO_OF_CHARS
    for i in string:
        count[ord(i)] += 1
    return count
 
# removeDirtyChars takes two
# string as arguments: First
# string (str)  is the one
# from where function removes dirty
# characters. Second  string
# is the string which contain all
# dirty characters which need
# to be removed  from first string
 
 
def removeDirtyChars(string, mask_string):
    count = getCharCountArray(mask_string)
    ip_ind = 0
    res_ind = 0
    temp = ''
    str_list = toList(string)
 
    while ip_ind != len(str_list):
        temp = str_list[ip_ind]
        if count[ord(temp)] == 0:
            str_list[res_ind] = str_list[ip_ind]
            res_ind += 1
        ip_ind += 1
 
    # After above step string is ngring.
     # Removing extra "iittg" after string
    return toString(str_list[0:res_ind])
 
 
# Driver code
mask_string = "mask"
string = "geeksforgeeks"
print removeDirtyChars(string, mask_string)
 
# This code is contributed by Bhavya Jain

C#

// C# program to remove
// duplicates, the order
// of characters is not
// maintained in this program
using System;
class GFG {
    static int NO_OF_CHARS = 256;
 
    /* Returns an array of size
    256 containing count of
    characters in the passed
    char array */
    static int[] getCharCountArray(String str)
    {
        int[] count = new int[NO_OF_CHARS];
        for (int i = 0; i < str.Length; i++)
            count[str[i]]++;
 
        return count;
    }
 
    /* removeDirtyChars takes two
    string as arguments: First
    string (str) is the one from
    where function removes dirty
    characters. Second string is
    the string which contain all
    dirty characters which need
    to be removed from first string */
    static String removeDirtyChars(String str,
                                   String mask_str)
    {
        int[] count = getCharCountArray(mask_str);
        int ip_ind = 0, res_ind = 0;
 
        char[] arr = str.ToCharArray();
 
        while (ip_ind != arr.Length)
        {
            char temp = arr[ip_ind];
            if (count[temp] == 0) {
                arr[res_ind] = arr[ip_ind];
                res_ind++;
            }
            ip_ind++;
        }
 
        str = new String(arr);
 
        /* After above step string
        is ngring. Removing extra
        "iittg" after string*/
        return str.Substring(0, res_ind);
    }
 
    // Driver Code
    public static void Main()
    {
        String str = "geeksforgeeks";
        String mask_str = "mask";
        Console.WriteLine(removeDirtyChars(str, mask_str));
    }
}
 
// This code is contributed by mits

Javascript

<script>
//Javascript Implementation
let NO_OF_CHARS  = 256;
function getcountarray(str2)
{
    var count = new Array(NO_OF_CHARS).fill(0);
  
    for (var i = 0; i < str2.length; i++)
    {
        count[str2.charCodeAt(i)]++;
    }
    return count;
}
  
 
/* removeDirtyChars takes two
string as arguments: First
string (str1)  is the one from
where function removes dirty
characters. Second  string(str2)
is the string which contain
all dirty characters which need
to be removed  from first
string */
function removeDirtyChars(str1, str2)
{
    // str2 is the string
    // which is to be removed
    var count = getcountarray(str2);
    var res ="";
       
    // ip_idx helps to keep
    // track of the first string
    var ip_idx = 0;
  
    while (ip_idx < str1.length)
    {
        var temp = str1[ip_idx];
        if (count[temp.charCodeAt(0)] == 0)
        {
            res = res.concat(temp);
        }
        ip_idx++;
    }
     
    return res;
}
  
// Driver Code
var mask_string = "mask"
var string = "geeksforgeeks"
document.write(removeDirtyChars(string, mask_string));
     
// This code is contributed by shivani
</script>

Output

geeforgee

Time Complexity: O(m+n) Where m is the length of the mask string and n is the length of the input string.

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