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

Last Updated : 13 Jul, 2023

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Given two strings string1 and string2, remove those characters from the first string(string1) which are present in the second string(string2). Both strings are different and contain only lowercase characters.
NOTE: The size of the first string is always greater than the size of the second string( |string1| > |string2|).

Example:

Input:
string1 = “computer”
string2 = “cat”
Output: “ompuer”
Explanation: After removing characters(c, a, t)
from string1 we get “ompuer”.

Input:
string1 = “occurrence”
string2 = “car”
Output: “ouene”
Explanation: After removing characters
(c, a, r) from string1 we get “ouene”.

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

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));
    }
}

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)

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;
}

C

#include <stdio.h>
#include <string.h>
 
char* removeChars(char string1[], char string2[]) {
    int i, j, k;
    int len1 = strlen(string1);
    int len2 = strlen(string2);
 
    for (i = 0; i < len2; i++) {
        for (j = 0; j < len1; j++) {
            if (string1[j] == string2[i]) {
                for (k = j; k < len1; k++) {
                    string1[k] = string1[k + 1];
                }
                len1--;
                j--;
            }
        }
    }
 
    return string1;
}
 
int main() {
    char string1[] = "geeksforgeeks";
    char string2[] = "mask";
    printf("%s\n", removeChars(string1, string2));
    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.

C#

// Include namespace system
using System;
public 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++)
    {
      var i = string2[index];
 
      // we find char exit or not
      while (string1.Contains(i.ToString() + ""))
      {
 
        // if char exit we simply remove that char
        string1 = string1.Replace((i.ToString() + ""),"");
      }
    }
    return string1;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    String string1;
    String string2;
    string1 = "geeksforgeeks";
    string2 = "mask";
    Console.WriteLine(GFG.removeChars(string1, string2));
  }
}
 
// This code is contributed by aadityaburujwale.

Javascript

// function to remove characters from string1 that are present in string2
function removeChars(string1, string2) {
    let i, j, k;
    let len1 = string1.length; // get length of string1
    let len2 = string2.length; // get length of string2
 
    // loop over characters in string2
    for (i = 0; i < len2; i++) {
        // loop over characters in string1
        for (j = 0; j < len1; j++) {
            // if character is found in both strings
            if (string1.charAt(j) == string2.charAt(i)) {
                // remove character from string1
                string1 = string1.substring(0, j) + string1.substring(j + 1);
                len1--; // decrease length of string1
                j--; // decrement j to account for shifted indices
            }
        }
    }
 
    return string1;
}
 
let string1 = "geeksforgeeks";
let string2 = "mask";
console.log(removeChars(string1, string2));

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: 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;
}

C

// C program to remove duplicates
#include <stdio.h>
#include <string.h>
 
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 = strlen(string1);
    int n2 = strlen(string2);
    printf("%s\n", removeChars(string1, n1, string2, n2));
    return 0;
}
 
//contributed by SATYAM NAYAK

Java

import java.util.*;
public class GFG {
    static String removeChars(String s1, int n1, String s2,
                              int n2)
    {
        String s3 = "";
 
        // an array of size 26 to count the frequency of
        // characters
        int[] arr = new int[26];
        for (int i = 0; i < 26; i++) {
            arr[i] = 0;
        }
        for (int i = 0; i < n2;
             i++) // assigned all the index of characters
            // which are present
            arr[s2.charAt(i) - 'a']
                = -1; // in second string by -1
        // (just flagging)
        for (int i = 0; i < n1; i++) {
            if (arr[s1.charAt(i) - 'a'] != -1) {
                // Checking if the index of
                // characters don't have -1
                s3 += s1.charAt(
                    i); // i.e, that character was not
                // present in second string and
                // then storing that character
                // in string
            }
        }
 
        s1 = s3;
        return s1;
    }
    // driver code
    public static void main(String args[])
    {
        String string1 = "geeksforgeeks";
        String string2 = "mask";
 
        int n1 = string1.length();
        int n2 = string2.length();
        System.out.println(
            removeChars(string1, n1, string2, n2));
    }
}
// This code is contributed by Samim Hossain
// Mondal.

Python3

# Python3 program to remove duplicates
def removeChars(s1, n1, s2, n2):
    s3 = ""
 
    # an array of size 26 to count the frequency of
    # characters
    arr = [0] * 26
 
    for i in range(0, n2):
        # assigned all the index of characters
        # which are present
        arr[ord(s2[i]) - ord('a')] = -1  # in second string by -1
    # (just flagging)
    for i in range(0, n1):
        if (arr[ord(s1[i]) - ord('a')] != -1):
            # Checking if the index of
            # characters don't have -1
            s3 += s1[i]  # i.e, that character was not
            # present in second string and
            # then storing that character
            # in string
 
    s1 = s3
    return s1
 
 
# driver code
string1 = "geeksforgeeks"
string2 = "mask"
n1 = len(string1)
n2 = len(string2)
print(removeChars(string1, n1, string2, n2))
 
# contributed by akashish__

C#

// C# program to remove duplicates
using System;
 
class GFG {
  static string removeChars(string s1, int n1, string s2,
                            int n2)
  {
    string s3 = "";
 
    // an array of size 26 to count the frequency of
    // characters
    int[] arr = new int[26];
    for (int i = 0; i < 26; i++) {
      arr[i] = 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
        s3 += s1[i]; // i.e, that character was not
        // present in second string and
        // then storing that character
        // in string
      }
    }
 
    s1 = s3;
    return s1;
  }
 
  // driver code
  public static void Main()
  {
    string string1 = "geeksforgeeks";
    string string2 = "mask";
    int n1 = string1.Length;
    int n2 = string2.Length;
    Console.WriteLine(
      removeChars(string1, n1, string2, n2));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

function removeChars(s1, n1, s2, n2) {
    let arr = new Array(26).fill(0); // an array of size 26 to count the frequency of characters
    let curr = 0;
    for (let i = 0; i < n2; i++) { // assigned all the index of characters which are present in second string by -1 (just flagging)
        arr[s2.charCodeAt(i) - 'a'.charCodeAt(0)] = -1;
    }
    let output = '';
    for (let i = 0; i < n1; i++) {
        if (arr[s1.charCodeAt(i) - 'a'.charCodeAt(0)] != -1) { // Checking if the index of characters don't have -1
            output += s1.charAt(i); // i.e, that character was not present in second string
        }
    }
    return output;
}
 
let string1 = "geeksforgeeks";
let string2 = "mask";
let n1 = string1.length;
let n2 = string2.length;
console.log(removeChars(string1, n1, string2, n2));

Output

geeforgee

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

Check if Strings Are Rotations of Each Other

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