Question about segment of code in C programming course - c

Question

I'm doing a course online in C and this is the assignment:
You are still conducting linguistic research! This time, you'd like to
write a program to find out how many letters occur multiple times in a
given word. Your program should read a word from the input and then
sort the letters of the word alphabetically (by their ASCII codes).
Next, your program should iterate through the letters of the word and
compare each letter with the one following it. If these equal each
other, you increase a counter by 1, making sure to then skip ahead far
enough so that letters that occur more than twice are not counted
again. You may assume that the word you read from the input has no
more than 50 letters, and that the word is all lowercase.
The solution they provide:
#include <stdio.h>
int main(void)
{
char word[51];
int length = 0;
int i, j;
char swap;
int repeats = 0;
scanf("%s", word);
while (word[length]!='\0')
length++;
//Sort the word by alphabetical order
for(j=0;j<length-1; j++) {
for(i=0;i<length-1;i++) {
if (word[i] > word[i+1]) {
swap = word[i];
word[i] = word[i+1];
word[i+1] = swap;
}
}
}
i = 0;
//Check for repeating characters in the sorted word
while (i<length-1) {
if (word[i]==word[i+1]) {
repeats++;
j=i+2;
//Continues through the word until it reaches a new character
while (j<length && word[i]==word[j])
j++;
i = j;
} else {
i++;
}
}
printf("%d", repeats);
return 0;
}
I understand everything up to code "//check for repeating characters in the sorted word//".
Specifically I don't understand the purpose or logic of "j=i+2" (especially the "+2") and how it relates to the next section of code "//continues through the word until it reaches a new character". I don't think this was adequately explained in the tutorials provided by the course.
Any insight or feedback is much appreciated.

Answer 1

Imagine your word is printed on paper, and you examine it through a small hole in a cardboard sheet. You can only see 2 letters at a time.
First, look at the beginning of the word. Identical letters? If not, shift by 1 position and repeat. If yes:
You found one repeated letter. Now you should find where the repeated run ends. To do that, shift your examination hole to the end of the repeated run.
It is possible to do this shift correctly in several ways. The way they implemented is until the first of the two visible letters is different from the repeated letters you found earlier. To do that, first you should shift by 2 positions, because at that position in code you know that the two letters are identical. But you could shift by 1 position instead — that would be correct too.
Another correct implementation — shift until you see two different letters in the hole. This may be easier to implement and more intuitive.