Finding duplicate chars in argv[]

Question

I'm trying to scan through my arguments list, argv[], and determine if there are duplicate chars. I've tried a few things, but nothing seems to work. I'm really new at C so bear with my horrible code:

char unavailableLetters[26] = "";
int pos = 0;
for(i = 0; i < (argc-1); i++) {
    int size = strlen(argv[i+1]);
    for(j = 0; j < size; j++) {
        char c = argv[i+1][j];

        if(j == 0) {
            unavailableLetters[pos] = c;
            pos+=1;
            } else {

            char *s = strchr (unavailableLetters, c);
            if(s == NULL) {
                unavailableLetters[pos] = c;
                pos += 1;
            }
        }
    }
}

My logic here is to parse through all the arguments and then parse through each character, first checking to see if it is contained in the array of unavailableLetters, and if it is not - add it, and move on. But no matter what I try, they either all get added, or none of them are added. This probably isn't the best way to do it, but I'm out of ideas.

1. List item

Answer 1

Here is a solution that I tested only briefly. It works with only lowercase letters, and exploits the fact that the ascii values of the letters are consecutive (ascii a is 97 which is why I subtract 97 when assigning an index and add it back when printing out the letter)

#include <iostream>
using namespace std;

int main (int argc, char *argv[])
{
    int letters[26];

    for (int i = 0; i < 26; i++) {
        letters[i] = 0;
    }

    for (int i = 1; i < argc; i++) {
        cout << "processing word: " << argv[i] << endl;
        char *word = argv[i];
        for (int j = 0; j < strlen(word); j++) {
           cout << "\tprocessing letter: " << word[j] << endl;
           letters[(int)word[j]-97]++;
        }
    }

    cout << "Letters that appear only once: " << endl;
    for (int i = 0; i < 26; i++) {
        if (letters[i] == 1)
            cout << (char) (i+97) << endl;
    }

    return 0;
}

Answer 2

Here is an alternate method. It uses a counter for each character, and then will display all characters the occur more than once:

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

int main(int argC, char *argV[])
   {
   int arg;
   char *cp;  
   int counters[256];

   /* Set all counters to zero. */
   memset(counters, 0, sizeof(counters));

   /* Iterate through each character of each argV */
   for(arg=1; arg<argC; ++arg)  // Change to "for(arg=0;..." to include argV[0]
      for(cp=argV[arg]; *cp; ++cp)
         ++counters[(int)*cp];  //Change to "++counters[toupper(*cp)];" to ignore case.

   /* Print characters that occurred more than once. */ 
   printf("Duplicate character list: ");   
   for(arg=0; arg<256; ++arg)
      if(counters[arg] > 1)
         printf("%c ", arg);

   printf("\n");   

   return 0;
   }

Answer 3

i wrote this on c++. This is more a pseudo code that a example. This work for the arguments with small letter only ok.

Your idea is good but this can work too. Try to maintain a array, that store in position i if the leter is found on the previous arguments. After that you can run over all the letters with a double for.

If when you hit a letter the value on the array is set to 1 the you already see that letter before and exist repeated letters. else set the letter position on the array to 1.

If the double for is entirely passed then any letter is repeated in the arguments.

Take a moment for thinking in this. This is the code

char unavalible_chars[26] = ""; // This is the array of ocurrences it only works with the small letters

    int i;
    for (i = 0; i < argc; i++)
    {
        char* c = argv[i];

        while ((char)*c) {

            int cPos = (*c) - 97; // Modifiy a little this for all letters

            if (unavalible_chars[cPos]){

                // Exist repeated
                return 0;
            }
            else
            {
                // Mark as present
                unavalible_chars[cPos] = 1;
            }
        }
    }

    return 1;