Question: Define an int function that removes all consecutive vowel repetitions from a string. The function should return the number of vowels removed and present the string without duplicates.
I am PT so Vogais is Vowels; Digite uma String is Write one String. A String sem duplicados fica assim ' %s ' e foram retiradas %d vogais is The string without duplicates is ' %s ' and where removed %d vowels.
Explanation: In portuguese we have some words with two consecutive vowels like: coordenador, coordenação (chqrlie example). But in thouse cases should be ignored in the context of this problem.
Problem: When I test a string like 'ooooo' it says the string without duplicate vogals is 'oo' and where removed 3 vowels. But it should be 'o' and 4 vowels removed. Another example with error is 'Estaa e umaa string coom duuuplicadoos', I am getting ' Esta e uma string com duplcdos ' and 8 vowels removed.
Note: This is a simple question so there isn't need to complicate. It only askes the consecutive duplicate vowels. The cases 'oOoO' -> 'oO' ,'abAb'->'abAb','abab' -> 'ab','aba'-> 'aba',... are in another chapter XD.
int Vogais(char *s) {
if (*s == 'A' || *s == 'a' || *s == 'E' || *s == 'e'
|| *s == 'I' || *s == 'i' || *s == 'O' || *s == 'o'
|| *s == 'U' || *s == 'u') return 1;
return 0;
}
int retiraVogaisRep(char *s) {
int res = 0;
for (int i = 0; i < strlen(s); i++) {
for (int j = i + 1; s[j] != '\0'; j++) {
if (s[i] == s[j] && Vogais(&s[j]) == 1) {
res++;
for (int k = j; s[k] != '\0'; k++) {
s[k] = s[k + 1];
}
}
}
}
return res;
}
int main() {
char s[38];
printf("Digite uma String:\n");
scanf("%[^\n]", s);
int res = retiraVogaisRep(s);
printf("A String sem duplicados fica assim ' %s ' e foram retiradas %d vogais.\n", s, res);
return 0;
}