C program to remove repeated char from a string

Question

I came across a interview question that asked to remove the repeated char from a given string, in-place. So if the input was "hi there" the output expected was "hi ter". It was also told to consider only alphabetic repititions and all the alphabets were lower case. I came up with the following program. I have comments to make my logic clear. But the program does not work as expectd for some inputs. If the input is "hii" it works, but if its "hi there" it fails. Please help.

#include <stdio.h>
int main() 
{
    char str[] = "programming is really cool"; // original string.
    char hash[26] = {0}; // hash table.
    int i,j; // loop counter.
// iterate through the input string char by char.
for(i=0,j=0;str[i];)
{
    // if the char is not hashed.
    if(!hash[str[i] - 'a'])
    {
        // hash it.
        hash[str[i] - 'a'] = 1;
        // copy the char at index i to index j.
        str[j++] = str[i++];
    }
    else
    {
        // move to next char of the original string.
        // do not increment j, so that later we can over-write the repeated char.
        i++;
    }
}

// add a null char.
str[j] = 0;

// print it.
printf("%s\n",str); // "progamin s ely c" expected.

return 0;

}

Answer 1

when str[i] is a non-alphabet, say a space and when you do:

hash[str[i] - 'a']

your program can blow.

ASCII value of space is 32 and that of a is 97 so you are effectively accessing array hash with a negative index.

To solve this you can ignore non-alphabets by doing :

if(! isalpha(str[i]) {
    str[j++] = str[i++]; // copy the char.
    continue;  // ignore rest of the loop.
}

Answer 2

这将打破任何空格字符（或'a'..'z'范围之外的任何其他内容）因为您正在超出哈希数组的边界进行访问。

Answer 3

这是代码高尔夫，对吗？

d(s){char*i=s,*o=s;for(;*i;++i)!memchr(s,*i,o-s)?*o++=*i:0;*o=0;}

Answer 4

void striprepeatedchars(char *str)
{
    int seen[UCHAR_MAX + 1];
    char *c, *n;

    memset(seen, 0, sizeof(seen));

    c = n = str;
    while (*n != '\0') {
        if (!isalpha(*n) || !seen[(unsigned char) *n]) {
            *c = *n;
            seen[(unsigned char) *n]++;
            c++;
        }
        n++;
    }
    *c = '\0';
}

Answer 5

...

// iterate through the input string char by char.
for(i=0,j=0;str[i];)
{
  if (str[i] == ' ')
  {
    str[j++] = str[i++];
    continue;
  }

    // if the char is not hashed.
    if(!hash[str[i] - 'a'])
    {

...

Answer 6

#include <stdio.h>
#include <string.h>

int hash[26] = {0};

static int in_valid_range (char c);
static int get_hash_code (char c);

static char * 
remove_repeated_char (char *s)
{
  size_t len = strlen (s);
  size_t i, j = 0;
  for (i = 0; i < len; ++i)
    {
      if (in_valid_range (s[i]))
    {
      int h = get_hash_code (s[i]);
      if (!hash[h])
        {
          s[j++] = s[i];
          hash[h] = 1;
        }
    }
      else
    {
      s[j++] = s[i];
    }
    }
  s[j] = 0;
  return s;
}

int
main (int argc, char **argv)
{
  printf ("%s\n", remove_repeated_char (argv[1]));
  return 0;
}

static int 
in_valid_range (char c)
{
  return (c >= 'a' && c <= 'z');
}

static int 
get_hash_code (char c)
{
  return (int) (c - 'a');
}

Answer 7

char *s;
int i = 0;

for (i = 0; s[i]; i++)
{
    int j;
    int gap = 0;
    for (j = i + 1; s[j]; j++)
    {
        if (gap > 0)
            s[j] = s[j + gap];
        if (!s[j])
            break;
        while (s[i] == s[j])
        {
            s[j] = s[j + gap + 1];
            gap++;
        }
    }
}