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".
Try It Yourself
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++ 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;
}
#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 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 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# 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
<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++ 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;
}
#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;
}
/*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.
# 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.
// 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.
// 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++ 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 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
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 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# 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.
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.
