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

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”

1. 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 */
2. 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
3. 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));
    }
}

Python3

# 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. 

Auxiliary Space: O(m)

Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.

An efficient solution is we find every character of string2 in string1 if that character is present then we simply erase that character from string1.

C++

// C++ program to remove duplicates
#include <bits/stdc++.h>
using namespace std;

string removeChars(string string1, string string2) {
       //we extract every character of string string 2
         for(auto i:string2)
        {
           //we find char exit or not
           while(find(string1.begin(),string1.end(),i)!=string1.end())
            {
                auto itr = find(string1.begin(),string1.end(),i);
               //if char exit we simply remove that char
                string1.erase(itr);
            }
        }
        return string1;
    }
// Driver Code
int main()
{
        string string1,string2;
        string1="geeksforgeeks";
        string2="mask";
         cout<<  removeChars(string1,string2)<<endl;;
       return 0;
}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static String removeChars(String string1,
                                     String string2)
    {
        // we extract every character of string string 2
        for (int index = 0; index < string2.length();
             index++) {
            char i = string2.charAt(index);
            // we find char exit or not
            while (string1.contains(i + "")) {
                int itr = string1.indexOf(i);
                // if char exit we simply remove that char
                string1 = string1.replace((i + ""), "");
            }
        }
        return string1;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String string1, string2;
        string1 = "geeksforgeeks";
        string2 = "mask";
        System.out.println(removeChars(string1, string2));
    }
}
//This code is contributed by KaaL-EL.

Python3

# Python 3 program to remove duplicates
def removeChars(string1, string2):

       # we extract every character of string string 2
    for i in string2:

        # we find char exit or not
        while i in string1:

            itr = string1.find(i)
           # if char exit we simply remove that char
            string1 = string1.replace(i, '')

    return string1

# Driver Code
if __name__ == "__main__":

    string1 = "geeksforgeeks"
    string2 = "mask"
    print(removeChars(string1, string2))

    # This code is contributed by ukasp.

Javascript

<script>

// JavaScript program to remove duplicates
function removeChars(string1, string2){

    // we extract every character of string string 2
    for(let i of string2){

        // we find char exit or not
        while(string1.indexOf(i) != -1){

            let itr = string1.indexOf(i)
            // if char exit we simply remove that char
            string1 = string1.replace(i, '')
        }
    }

    return string1
}

// Driver Code
let string1 = "geeksforgeeks"
let string2 = "mask"
document.write(removeChars(string1, string2))

// This code is contributed by shinjanpatra

</script>

Output

geeforgee

Time Complexity: O(n*m), where n is the size of given string2 and m is the size of string1.
Auxiliary Space: O(1), as no extra space is used

Efficient Solution – O(|s1|) time complexity and O(1) space complexity.

An efficient solution is that we can mark the occurrence of all characters present in second string by -1 in frequency character array and then while traversing first string we can ignore the marked characters as shown in below program. 

C++

// C++ program to remove duplicates
#include <bits/stdc++.h>
using namespace std;

char* removeChars(char* s1, int n1, char* s2, int n2)
{
    int arr[26] = { 0 }; // an array of size 26 to count the frequency of characters
    int curr = 0;
    for (int i = 0; i < n2; i++) // assigned all the index of characters which are present
        arr[s2[i] - 'a'] = -1;   // in second string by -1 (just flagging)
    for (int i = 0; i < n1; i++)
        if (arr[s1[i] - 'a'] != -1) {  // Checking if the index of characters don't have -1
            s1[curr] = s1[i];       // i.e, that character was not present in second string
            curr++;               // and then storing that character in string
        }
    s1[curr] = '\0';    // marking last character as null to point the end of string
    return s1;
}

// driver code
int main()
{
    char string1[] = "geeksforgeeks";
    char string2[] = "mask";
    int n1 = sizeof(string1) / sizeof(string1[0]);
    int n2 = sizeof(string2) / sizeof(string2[0]);
    cout << removeChars(string1, n1, string2, n2) << endl;
    return 0;
}

Output:

geeforgee

Time Complexity: O(|S1|), where |S1| is the size of given string 1.
Auxiliary Space: O(1), as only array of constant size (26) is used. 

Thank you!