Sorting word occurrence in string C programming - c

Question

I have been stuck on this for a while now. I wrote my program to count word occurrence in an inputted string by the user as well to sort the words alphabetically. My issue is my program runs perfectly only if there are spaces in between the words inputted. For example, if I input "to to," my program will count those two words as two different words due to the comma rather than counting it as one word in "to" as I would like it to. It is that issue for all of my delimiters in the array const char delim[]. How can I fix this issue in my program? I really appreciate any help! My code is down below:
Edit: I took Bob's suggestion to use strchr() and it worked! My only issue is my program outputs the count for delimiters now. I was thinking of possibly writing an if statement when comparing words[i] with words[j] to see if they have the same value. Is that the best approach to it?
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
const char delim[] = ", . - !*()&^%$##<> ? []{}\\ / \"";
#define SIZE 1000
int main(){
char string[SIZE], words[SIZE][SIZE], temp[SIZE];
int a = 0, i = 0, j = 0, k = 0, n = 0, count;
int c = 0, cnt[26] = { 0 };
int word = 0;
int x;
printf("Enter your input string:");
fgets(string, SIZE, stdin);
string[strlen(string) - 1] = '\0';
lower(string);
/*extracting each and every string and copying to a different place */
while (string[i] != '\0'){
if (strchr(", . - !*()&^%$##<> ? []{}\\ / \"", string[i]) != NULL){
words[j][k] = '\0';
k = 0;
j++;
}else {
words[j][k++] = string[i];
}
i++;
}
words[j][k] = '\0';
n = j;
printf("Number of occurences of each word unsorted:\n");
i = 0;
/* find the frequency of each word and print the results */
while (i <= n) {
count = 1;
if (i != n) {
for (j = i + 1; j <= n; j++) {
if (strcmp(words[i], words[j]) == 0) {
count++;
for (a = j; a <= n; a++)
strcpy(words[a], words[a + 1]);
n--;
}
}//for
}
//word == strtok(string, delim);
/* count - indicates the frequecy of word[i] */
printf("%s\t%d\n", words[i], count);
i = i + 1;
}//while
printf("Alphabetical Order:\n");
/* sort the words in the given string */
for (i = 0; i < n; i++){
strcpy(temp, words[i]);
for (j = i + 1; j <= n; j++){
if (strcmp(words[i], words[j]) > 0){
strcpy(temp, words[j]);
strcpy(words[j], words[i]);
strcpy(words[i], temp);
}
} //inner for
} //outer for
i = 0;
while (i <= n) {
count = 1;
if (i != n) {
for (j = i + 1; j <= n; j++) {
if (strcmp(words[i], words[j]) == 0) {
count++;
}
}
}
printf("%s\n", words[i]);
i = i + count;
}
}

Answer 1

Strip every word of that delimeter before comparing. Actually you don't even need a list of delimeters because words are 'alpha' other than that it's a delimeter.

Answer 2

Please try this, it works, it is an extract of your own code, a little bit modified, it will give you the count, then you have to write the rest, have fun.
#include <string.h>
#define YES 1
#define NO 0
int main( )
{
char string[1000];
int i = 0;
int j = 0;
int is_this_a_word = 0;
strcpy( string, " to or not ,tobe" );
while ( string[i++] != '\0' )
{
if ( strchr( ", . - !*()&^%$##<> ? []{}\\ / \"", string[i] ) != NULL )
{
is_this_a_word = NO;
}
else
{
if ( ! is_this_a_word )
{
is_this_a_word = YES;
j++;
}
}
}
printf( "I counted %d words", j );
getchar( );
}