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Remove characters from the first string which are present in the second string
  • Difficulty Level : Easy
  • Last Updated : 23 Apr, 2021

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

We strongly recommend that you click here and practice it, before moving on to the solution.

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

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

Implementations: 

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;
}

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;
}

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
Output
geeforgee

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

 

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