Created
September 14, 2018 04:28
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Vowels Sample answer
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#include<stdio.h> | |
#include<string.h> | |
#define TRUE 1 | |
#define FALSE 0 | |
#define LENGTH_LIMIT 1000 | |
int is_char_vowel(char c) { | |
switch (c) { | |
case 'a': | |
case 'e': | |
case 'i': | |
case 'o': | |
case 'u': | |
return TRUE; | |
default: | |
return FALSE; | |
} | |
} | |
int main(int argc, char **argv) { | |
int i; | |
char current_char; | |
char *vowel_order = "aeiou"; | |
char last_vowel; | |
int last_vowel_idx = -1; | |
int is_correct; | |
for (i = 0, is_correct = TRUE; i < LENGTH_LIMIT; i++) { | |
current_char = argv[1][i]; | |
if (current_char == '\0') { | |
break; | |
} | |
if (current_char == last_vowel) { | |
continue; | |
} | |
if (is_char_vowel(current_char)) { | |
if ( | |
last_vowel_idx != -1 && | |
current_char != vowel_order[last_vowel_idx + 1] | |
) { | |
is_correct = FALSE; | |
break; | |
} | |
last_vowel = current_char; | |
last_vowel_idx = last_vowel_idx == -1 ? 0 : last_vowel_idx + 1; | |
} | |
} | |
if (!is_correct || last_vowel_idx != 4) { | |
printf("False\n"); | |
return 0; | |
} | |
printf("True\n"); | |
return 0; | |
} |
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Given a word, check if it has all vowels in order (a, e, i, o, u). The words are only provided in lower case.
Example:
facetious => true
precarious => false
apple => false