End Google Ads 201810 - BS.net 01 --> hi, i have C exam next week, so i started solving questions from previous years exams and run into this question, the answer i come up to is far far from having reasonable length, it's too complicated and long , but i can't come up anything shorter.

the question is: write a function that receives an array of letters , and a pointer to final result string that hasn't allocated yet, the function supposed to take the array of letters and count the appearance of every letter and make the final result string look like that:

received array of letters : "abbacdeefg"
result string: "a 2;b 2;c 1;d 1;e 2;f 1;g 1"
in the result it should be alphabetically printed

here is the code that i wrote, it just doesn't makes sense that this is the actual solution:

void sortarray(char str[],char * final) { int i = 1,j, extra = 0; char * letters, *lerrptr; int * count, * cerrptr; int size = 0, flag = 0; // allocate memory for one array storing letter and one for storing it's appearance letters = (char *) malloc(sizeof(char) * 1); if(letters == NULL) exit(1); count = (int *) malloc(sizeof(int) * 1); if(count == NULL) exit(1); // fill the first letter to array letters[0] = str[0]; count[0] = 1; size++; // scan the whole string while(str[i] != 0) { // initiate flag for new run flag = 0; for(j = 0 ; j < size ; j++) { // if the letter already appeared before just increment the count and raise the flag that no new allocating is necessary if(str[i] == letters[j]) { count[j]++; flag = 1; } } // if the flag of already appeared hasn't been raised make new al******** for new letter if(!flag) { flag = 0 ; size++; lerrptr = letters; letters = (char *) realloc(letters,sizeof(char) * size); if(letters == NULL) { free(lerrptr); free(count); exit(1); } cerrptr = count; count = (int *) realloc(count,sizeof(int) * size); if(letters == NULL) { free(cerrptr); free(letters); exit(1); } // insert new letter and 1 for the count letters[size-1] = str[i] ; count[size-1] = 1 ; } i++; } // bubble sort the letters for(i = 0 ; i < size - 1; i++) { for(j = 0 ; j < size - 1 - i ; j++) { if(letters[j] < letters[j - 1]) { letters[j] ^= letters[j+1]; letters[j+1] ^= letters[j]; letters[j] ^= letters[j+1]; count[j] ^= count[j+1]; count[j+1] ^= count[j]; count[j] ^= count[j+1]; } } } // check for more 9 appearances to reserve more letters for one more digit for(i = 0 ; i < size ; i++) if(count[i] > 9) extra++; // allocate the memory for the final array that is going to be printed final = (char *) malloc(sizeof(char) * (size * 4) + extra + 1); j = 0; for(i = 0 ; i < size ; i++) { // insert the letter final[i*4 + j] = letters[i]; // insert space final[i*4 + 1 + j] = ' '; // if more then one digit used for the count of appearances if(count[i] > 9) { // first digit final[i*4 + 2 + j] = count[i]/10 + 48; j++; // second final[i*4 + 2 + j] = count[i]%10 + 48 ; } else final[i*4 + 2 + j] = count[i] + 48; // print ; final[i*4 + 3 + j] = ';'; } // add terminating character final[i*4 + j] = 0; // print the result printf(" %s ", final); } can anyone help me come up with more reasonable solution?