Program to count length of each word in string in C - c

Question

I'm writting a program to count the length of each word in array of characters. I was wondering if You guys could help me, because I'm struggling with it for at least two hours for now and i don't know how to do it properly.
It should go like that:
(number of letters) - (number of words with this many letters)
2 - 1
3 - 4
5 - 1
etc.
char tab[1000];
int k = 0, x = 0;
printf("Enter text: ");
fgets(tab, 1000, stdin);
for (int i = 2; i < (int)strlen(tab); i++)
{
for (int j = 0; j < (int)strlen(tab); j++)
{
if (tab[j] == '\0' || tab[j]=='\n')
break;
if (tab[j] == ' ')
k = 0;
else k++;
if (k == i)
{
x++;
k = 0;
}
}
if (x != 0)
{
printf("%d - %d\n", i, x);
x = 0;
k = 0;
}
}
return 0;

Answer 1

By using two for loops, you're doing len**2 character scans. (e.g.) For a buffer of length 1000, instead of 1000 character comparisons, you're doing 1,000,000 comparisons.
This can be done in a single for loop if we use a word length histogram array.
The basic algorithm is the same as your inner loop.
When we have a non-space character, we increment a current length value. When we see a space, we increment the histogram cell (indexed by the length value) by 1. We then set the length value to 0.
Here's some code that works:
#include <stdio.h>
int
main(void)
{
int hist[100] = { 0 };
char buf[1000];
char *bp;
int chr;
int curlen = 0;
printf("Enter text: ");
fflush(stdout);
fgets(buf,sizeof(buf),stdin);
bp = buf;
for (chr = *bp++; chr != 0; chr = *bp++) {
if (chr == '\n')
break;
// end of word -- increment the histogram cell
if (chr == ' ') {
hist[curlen] += 1;
curlen = 0;
}
// got an alpha char -- increment the length of the word
else
curlen += 1;
}
// catch the final word on the line
hist[curlen] += 1;
for (curlen = 1; curlen < sizeof(hist) / sizeof(hist[0]); ++curlen) {
int count = hist[curlen];
if (count > 0)
printf("%d - %d\n",curlen,count);
}
return 0;
}
UPDATE:
and i don't really understand pointers. Is there any simpler method to do this?
Pointers are a very important [essential] tool in the C arsenal, so I hope you get to them soon.
However, it is easy enough to convert the for loop (Removing the char *bp; and bp = buf;):
Change:
for (chr = *bp++; chr != 0; chr = *bp++) {
Into:
for (int bufidx = 0; ; ++bufidx) {
chr = buf[bufidx];
if (chr == 0)
break;
The rest of the for loop remains the same.
Here's another loop [but, without optimization by the compiler] double fetches the char:
for (int bufidx = 0; buf[bufidx] != 0; ++bufidx) {
chr = buf[bufidx];
Here is a single line version. Note this is not recommended practice because of the embedded assignment of chr inside the loop condition clause, but is for illustration purposes:
for (int bufidx = 0; (chr = buf[bufidx]) != 0; ++bufidx) {