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 first string which are present in second string (mask string).

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

4: Put a ‘\0′ at the end of the string?

Implementations:

• C
• Python

#include <stdio.h> #include <stdlib.h> #define NO_OF_CHARS 256   /* Returns an array of size 256 containg 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 program to test getCharCountArray*/ int main() {     char str[]         = "geeksforgeeks";     char mask_str[]  = "mask";     printf("%s", removeDirtyChars(str, mask_str));     return 0; }

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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