Interview Questions And Answers 
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You are given two strings: S1 and S2. Delete from S2 all those characters which occur in S1 also and create a clean S2 with the relevant characters deleted.
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#include<stdio.h>
#include<string.h>
main() {
char str1[5] = “abcd”;
char str2[3] = “bc”;
int len = strlen(str1);
int i =0;
char newstr[len];
int cntr = 0;
for ( i = 0; i<len; i++ ) {
if ( strchr(str2,str1[i]) == NULL ) {
newstr[cntr++] = str1[i];
}
}
newstr[cntr] = ‘ ’;
printf(”%s%s%s”, ” the new str is “, newstr, “n”);
}
char* ClearString (char* s1, char* s2)
{
unsigned int count;
unsigned char* str;
unsigned chat* insert;
unsigned char map[32];
if(s1==NULL
- in first iteration, insert all the values of s1 to a hash table say T.
- in second iteration, for s2 hash every character to T, if found make this location NULL and if found move this character to adjacent location if adjacent locaton is NULL
A linear solution without using a hash table.
void substring_remove(string &str, string &substr)
{
int i=0;
int len=str.length();
while(i=0)
{
i=str.find_first_of(substr,i);
if(i>=0)
str.erase(i,1);
}
}
int len = strlen(a);
int index = 0;
void ClearString(char *a,char *b)
{
int len = strlen(a);
int index = 0;
for(int i=0;i
int deleteChars(char* s1,char* s2)
{
int arr[256];
for(int i=0;i
char* s1minuss2(char *s1, char* s2)
{
static char res[20]= “”;
int i = 0, j = 0;
while ( s2[i] != ” )
{
if ( !strchr(s1, s2[i]) )
{ res[j] = s2[i]; j++; }
i++;
}
res[j] = ”;
return res;
}
O(n) approach. will remove any occurance of the characters.
# include
# include
# include
int table [ 128 ] ;
void main()
{
char *str=”this is a test”;
char *str1=”is “;
clrscr();
while(*str1!=”)
{
table[toupper(*str1)]++;
str1++;
}
while(*str!=”)
{
if(table[toupper(*str)]==0)
{
printf(”%c”,*str);
}
str++;
}
getch();
}
void Delete(char* string1, char* string2)
{
char* s1 = string1, s2 = string2;
int index, destIndex = 0;
int* lookUp = ((int *)(calloc(255, sizeOf(int)));
while(” != *s2)
{
int index = *s2;
lookUp[index]++;
s2++;
}
while(” != *s1)
{
index = *s1;
if(0
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