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Print non used character in alphabet

Time:11-04

I'm trying to create a program in C that prints the letters from alphabet that are not in gussed_character. All unused letters are saved in available_character.

Code:

#include <stdio.h>
#include <ctype.h>
#include <string.h>

void letters(const char gussed_character[], char available_character[]){
    char alphabet[] = "abcdefghijklmnopqrstuvwxyz";
    int gussed_character_lenght = strlen(gussed_character);
    int alphabet_lenght = strlen(alphabet);
    int counter = 0;

    for(int i = 0; i<alphabet_lenght; i  ){
        counter = 0;
        for(int j = 0; j<gussed_character_lenght; j  ){
            if(alphabet[i] != gussed_character[j]){
                counter  ;                              
            }
            if(counter == gussed_character_lenght){
                available_character[i] = alphabet[i];
            }           
        }
    }
    printf("%s", available_character);
}

int main(){
    char result[30];
    letters("arpstxgoieyu", result);  
}

Example from code:

letters("arpstxgoieyu", result);

This letters are used and prgram should print this letters:

bcdfhjklmnqvwz

But my program print this:

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