Repeated characters in a string, C program

You are thinking too hard. :-) Here is help to think it through. You only need a single index into the string, plus two pieces of information: current character and count.

So, given the string “RRDDuuuRRRuuu”:

Start with the first character, R. That is your current character.

Loop through the string, counting the number of times you find R.

RRDDuuuRRRuuu
^
current = 'R'
count   = 1

RRDDuuuRRRuuu
 ^
current = 'R'
count   = 2

RRDDuuuRRRuuu
  ^
oh no! we have found a DIFFERENT character!

At this point, we know we found two Rs. Print it.
Then set the current character to the new character (D) and reset your count to one, then let the loop continue as before.

RRDDuuuRRRuuu
  ^
current = 'D'
count   = 1

RRDDuuuRRRuuu
   ^
current = 'D'
count   = 2

RRDDuuuRRRuuu
    ^
new character again

Stop when the new current character would be a \0.

RRDDuuuRRRuuu
             ^
um, new character would be '\0', so after printing the “3u” we are done.

Good luck!

EDIT: Well, since everyone is just posting solutions, once you have done your own you can look at the code for mine.

#include <stdio.h>

void solution(const char * s)
{
  char current = s[0];
  int  count = 1;
  while (*s++)
  {
    if (*s == current)
    {
      count += 1;
    }
    else
    {
      printf("%d%c ", count, current);
      current = *s;
      count = 1;
    }
  }
  puts("");
}

int main(int argc, char ** argv)
{
  if (argc == 2)
    solution(argv[1]);
  return 0;
}

Linux:

$ clang -Wall -Wextra -Werror -pedantic-errors -O3 a.c
$ ./a.out RRDDuuuRRRuuu
2R 2D 3u 3R 3u
$ ./a.out abb
1a 2b
$ ./a.out aab
2a 1b
$ ./a.out a
1a
$ ./a.out ""

$ ./a.out
$ █

Windows:

C:> cl /nologo /EHsc /Ox a.c
a.c
C:> a.exe RRDDuuuRRRuuu
2R 2D 3u 3R 3u
C:> a.exe abb
1a 2b
C:> a.exe aab
2a 1b
C:> a.exe a
1a
C:> a.exe ""

C:> a.exe
C:> █

Your program only handles the first 2 characters, assuming there are at least 2 sets of characters. It will have undefined behavior if the string only has a single character, repeated or not.

You should wrap the counting loop in an outer loop that repeats until the end of the string.

Note also these problems:

• tell scanf the maximum number of characters to store into the destination array to avoid a buffer overflow on longer input

• check the return value of scanf() to detect invalid or missing input.

• always append a newline at the end of your output.

• j = i - 1; is cumbersome and works by coincidence. j = i is more consistent with your do/while loop logic.

  IMHO you should avoid do/while loop completely. These constructions are notoriously error prone and lead to confusing code, and many logical bugs. Use for or while loops instead.

Here is a modified version:

#include <stdio.h>

int main(void)
{
    char buf[3000];  // Declaring string arrays
    
    printf("Enter a string: ");
    fflush(stdout);  // ensure the prompt is visible to the user
    if (scanf("%2999s", buf) != 1) {
        printf("input error\n");
        return 1;
    }

    const char *p = buf;
    while (*p) {
        int count = 1;
        char ch = *p++;
        while (*p == ch) {
            count++;
            p++;
        }
        printf("%d%c ", count, ch);
    }
    putchar('\n');
    return 0;
}

The only two vars you need - char counter and pointer for iterations or index (with pointer less text).

There's enough one loop with comparing current char with next and exit loop on '\0' char;

#include <stdio.h>

int main() {
    char str1[3000];
    char *c = str1;
    printf("Enter a string: ");
    if(scanf("%2999s", str1) != 1) return 1; //Protect from overflow and empty
    for (int n = 1; *c; c++,n++)
        if (*c != *(c+1)){
            printf("%d%c ", n, *c);
            n=0;
        }
    return 0;
}

https://godbolt.org/z/nq1z1qqMa