C program to remove all repeated characters in a string

Question

I want to remove all the repeated characters from a string. For example, if I have:

"aapuroo"

I want the result to be

apuro

but it shows

apuroo

what is my mistake here?I am a beginer pardon my mistakes

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

int main()
{
    char str[100];
    int i,j,k,n;
    gets(str);

    n = strlen(str);

    for(i=0; i<n; i++)
    {

        for(j=i+1; j<n;j++)
        {
             if(str[i] == str[j])
             {
                 str[j] = str[j+1];
             }

        }
    }

    puts(str);

    return 0;
}

Working with a result buffer rather than in-place will make it way simpler. — Eugene Sh., Dec 23 '19 at 21:08
Also don't use `gets`: https://stackoverflow.com/questions/1694036/why-is-the-gets-function-so-dangerous-that-it-should-not-be-used — UnholySheep, Dec 23 '19 at 21:26
Note that it is never safe to use `gets()` — see [Why `gets()` is too dangerous to be used — ever!](https://stackoverflow.com/questions/1694036/why-is-the-gets-function-so-dangerous-that-it-should-not-be-used) for the sordid details. — Jonathan Leffler, Dec 23 '19 at 21:27
What do you want as the output from the input `abracadabra`? Two main options are `abracadabra` (because no two adjacent characters are the same) or `abrcd` because all the letters except `d` are repeated. — Jonathan Leffler, Dec 23 '19 at 21:31

Vlad from Moscow · Accepted Answer · 2019-12-23T21:45:41.580

The approach is invalid. Consider a string like char str[] = "111";.

So in these loops

for(i=0; i<n; i++)
{

    for(j=i+1; j<n;j++)
    {
         if(str[i] == str[j])
         {
             str[j] = str[j+1];
         }

    }
}

str[0] is equal tp '1'. str[1] is also equal to '1'. So str[1] is substituted for str[2] that is also equal tp '1'. As a result you will get that the array contains the string "11"

Moreover the inner loop processes not only adjacent repeated characters but all duplicated characters in the string.

Pay attention to that the function gets is an unsafe function and is not supported by the C Standard any more. Use instead the standard function fgets.

The program can look the following way

#include <stdio.h>

int main()
{
    enum { N = 100 };
    char str[N];

    fgets( str, N, stdin );

    for ( size_t i = 0, j = 0;  str[i] != '\0'; )
    {
        if ( str[++i] != str[j] && str[i] != '\n' ) 
        {
            if ( i != ++j ) str[j] = str[i];
        }           
    }

    puts( str );

    return 0;
}

If to enter the string "aapuroo" then the program output will be

apuro

The subexpression str[i] != '\n' used in the if statement is present because the function fgets can append the new line character to the entered string. So this subexpression is used to exclude the new line character from the result string. Or you could before the loop to remove this character.

@MustabirApurbo Try not to use "magic numbers". Use named constants. — Vlad from Moscow, Dec 23 '19 at 21:51

score 0 · Answer 2 · answered Dec 23 '19 at 21:32

When you find a repeated character, you need to shift the rest of the string to the left, filling the last byte with a zero. As for your code, you just duplicate the [j+1] character and then, in your case, exiting since you are at the end of the string.

score -1 · Answer 3 · edited Dec 24 '19 at 02:34

-1

#include <stdio.h>

int main()
{
  char str[100];
  // this programm is created by chandan pradhan
  gets(str);
  for ( size_t i = 0, j = 0;  str[i] != NULL;)
  {
    if ( str[++i] != str[j] && str[i] != '\n' ) 
    {
      if ( i != ++j ) 
        str[j] = str[i];
    }
  }
  puts( str );
  return 0;
}

edited Dec 24 '19 at 02:34

chux - Reinstate Monica

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answered Dec 24 '19 at 01:59

Chandan Pradhan

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C program to remove all repeated characters in a string

3 Answers3