CareerCup

there are many ways to do it
if memory is a don't care
then we can simply use indication array :)

string ans="";
bool used[256]; // for ascii code
memset(used, false, sizeof(used));
for(int i=0; i<str.length(); ++i) 
   if(used[str[i]]) ;
   else {
          used[str[i]]=true;
          ans+=str[i];   
       }

hi,

it seems like you are adding a character to the answer string immediately after you find that it hasn't been found before. But what if a character is repeated sometime later and now it has already been added to the answer string. Aren't you supposed to remove it from the answer string. So i think this solution is wrong.

Please see this,

char str[50]="I am an Indian, which country you belong to?\0";
int i;
for(i=0; '\0' !=str[i]; i++)
{
int j;
for(j=i+1; '\0' !=str[j]; j++)
{
if(str[i] == str[j])
{
int k;
for(k=j;str[k]!='\0';k++)
{
str[k]=str[k+1];
}
}
}
}

The output is "I amndi,whcoutrybelg?", I think it is expected... :)
Please excuse if there is anything wrong in the sentence

@Rajesh...go jump...cant u see the better solution above..u algo takes O(n2) + times to remove the duplicates..

How about this soln.
Read each character one by one and put it in hashtable.Here key will be alphabet itself and value will be its repetition.whenever you get the count more than 1 in hashtable delete the aplhablet from string.

Reading each character and putting in hashtable.O(n) + const time to put in hashtable.

@sourab thts wat the above soln does

void RemoveDups(char *str)
{
struct hastable *newhash=createHashtable();
int current=0,new=0;
char *str1=malloc(sizeof(str));
while(str[current]!=NULL)
{
If (!Hashpresent(newhash,str[current]))
{
Hashinsert(newhash,str[current]);
str1[new]=str[current];
new=new+1;
}
current=current+1;
}
str1[current]='\0';
}

This will solve purpose in fast and simple code.
It gives result in O(n)

void removeduplicate(char *str)
{
int checker = 0;
int cnt=0;
for(int i=0;i<strlen(str);i++)
{
int val =*(str+i)- (int)'a';
if ((checker & (1 << val)) > 0)
continue;
else{
*(str+cnt)=*(str+i);
cnt++;
}
checker |= (1 << val);
}
*(str+cnt)='\0';
}

//c# O(n)
        static char[] removeDuplicateChars(char[] _str)
        {
            bool[] charset = new bool[256];
            for (int i = 0; i < _str.Length; i++)
            {
                int val = _str[i];
                if (charset[val]) { _str[i] = ' '; }
                charset[val] = true;
            }
            return _str;
        }