C Programming, what am I doing wrong? Strtok - c

Question

In the notepad that is saved in the same file as my main.c program is,
ape apple
ball bill bull
foot
parrot peeble
season
zebras zoo
For example, assume that the word “bull” is in the dictionary. The word “bull” contains 1 ‘b’ character, 2 ‘l’ characters, and 1 ‘u’ character. Now say the input letters were “alblldi”. In “alblldi”, we have enough ‘b’ characters for “bull”, since “alblldi” contains at least 1 ‘b’ character. Similarily, “alblldi”, has enough ‘l’ characters for “bull”, since “alblldi” contains at least 2 ‘l’ characters. However, “alblldi” does not have at least 1 ‘u’ character, and as such we know that we cannot make “bull” from “alblldi”.
How do I achieve this?
So I just started to type this code and I need help, so far I got:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SIZE 100
int main( void )
{
int found = 0;
char string[SIZE];
char name[ SIZE ];
FILE *cfPtr;
char word[ SIZE ];
char *tokenPtr; // create char pointer
printf("\nGive me a sentence: ");
fgets( string, SIZE, stdin );
printf("The string to be tokenized is: %s\n", string);
printf("Give me a word: ");
scanf("%s",word);
tokenPtr = strtok( string, "" ); // begin tokenizing sentence
puts("");
// continue tokenizing sentence until tokenPtr becomes NULL
while ( tokenPtr != NULL ) {
if (!strcmp(word,tokenPtr)) {
printf("%s : This is the word you are looking for!\n", tokenPtr);
found = 1;
}
else {
printf( "%s\n", tokenPtr );
}
tokenPtr = strtok( NULL, " " ); // get next token
} // end while
if (!found) {
printf("The word \"%s\" was not found in the sentence\n",word);
}
if ( ( cfPtr = fopen( "dictionary.txt", "r" ) ) == NULL ) {
puts( "File could not be opened" );
}
else {
fgets( name, SIZE, cfPtr );
while ( !feof( cfPtr ) ) {
printf( "Line from file is: %s\n", name );
tokenPtr = strtok( name, "\t" ); // begin tokenizing sentence
// continue tokenizing sentence until tokenPtr becomes NULL
while ( tokenPtr != NULL ) {
printf( "\t%s\n", tokenPtr );
tokenPtr = strtok( NULL, "\t" ); // get next token
} // end while
fgets( name, SIZE, cfPtr );
}
fclose( cfPtr );
}
}

Answer 1

Obviously homework so I'm not doing it for you, but: You can do it on paper well enough to ask a clear question, so just figure out how to implement that in code.
A counter for each letter `int a=0; int b=0; int z=0;... (one variable per letter? there has to be a better way...)
set your counters based on the input word
check if the word you are searching can satisfy all of the counters
There are a couple of easily identified common routines here.

Answer 2

I would do it like so:
#include <string.h>
#include <stdio.h>
int count(char *str, int c) {
int n = 0;
for (size_t i = 0; i < strlen(str); i++) {
if (str[i] == c) {
n++;
}
}
return n;
}
int check_word(int counts[256], char *str) {
if (str == NULL) {
return 0;
}
for (int i = 0; i < 256; i++) {
if (count(str, i) < counts[i]) {
return 0;
}
}
return 1;
}
int check_words(char *word, char **words, size_t n) {
int counts[256] = {0};
for (size_t i = 0; i < strlen(word); i++) {
counts[(int)word[i]]++;
}
int k = 0;
for (size_t i = 0; i < n; i++) {
if (check_word(counts, words[i])) {
printf(">>> %s\n", words[i]);
k++;
} else {
printf(" %s\n", words[i]);
}
}
return k;
}
int main(void) {
char *words[] = {
"bull", "bill", "ball", "bubblel", "babble", "bible",
};
check_words("bull", words, 6);
return 0;
}
I initially build an array containing the count for each character in the initial word, and then for each string in the array, I check whether it contains enough of each.
Running this code gives me the following output, which I expected.
>>> bull
bill
ball
>>> bubblel
babble
bible
meaning that bull matches bull, and bubblel.