How to check for repeated characters within a string in c

Here are the basics of a solution:

When starting, initialize a table of flags: char Seen[UCHAR_MAX+1] = {0};. In this array, Seen[c] will be true (non-zero) if and only if character c has been seen already. (To get UCHAR_MAX, #include <limits.h>.)

When processing each character, copy it to an unsigned char: unsigned char c = argv[1][i];. Then convert it to uppercase: c = toupper(c);. Then test whether it has been seen already:

if (Seen[c])
    Report error...

If it is new, remember that it has been seen: Seen[c] = 1;.

That is all that is necessary.

Notes:

• If it is known that A to Z are contiguous in the character set, as they are in ASCII, then char Seen[UCHAR_MAX+1] can be reduced to char Seen['Z'-'A'+1] and indexed using Seen[c-'A']. (Actually, a weaker condition suffices: A is the lowest uppercase character in value and Z is the greatest.)
• Do not be tempted to use unsigned char c = toupper(argv[1][i]), because toupper is defined for unsigned char values, and a char value may be out of bounds (negative).
• In an esoteric C implementation where a char is as wide as an int, none of which are known to exist, UCHAR_MAX+1 would evaluate to zero. However, the compiler should warn if this happens.

I'm assuming this was from cs50, I'm on this problem right now... What I did was I used a simple boolean function. I used two loops, the first loop was to represent the first character of reference and inside the nested loop I looped through all the other chars and returned false if i == k.

bool no_repeat(string arg_letters)
{
   for (int i = 0, j = strlen(arg_letters); i < j; i++)
   {
      for (int k = i+1; k < j; k++)
      {
         if (arg_letters[i] == arg_letters[k])
         {
            return false;
         }

      }
   }
   return true;
}

From the question, the expected input is 26 characters with each appears only 1 time.

According to Andreas Wenzel's idea, i add counters for each character. in case any character missing or duplicate will be spotted. Here's my solution:

int main(int argc, char *argv[])
{
    const int OFFSET = 'A';

    if (argc != 2)
    {
        return 1;
    }
    if (strlen(argv[1]) != 26)
    {
        printf("Key must contain 26 characters.\n");
        return 1;
    }

    unsigned char *key = argv[1];
    int count[26] = {0}; // array of all zero

    // Check for any invalid character 
    for (int i = 0, n = strlen(key); i < n; i++)
    {
        if (!isalpha(key[i]))
        {
            printf("Key contains invalid characters.\n");
            return 1;
        }
        count[toupper(key[i]) - OFFSET]++;
    }
    // check for duplicate character of 
    for (int i = 0; i < 26; i++)
    {
        if (count[i] != 1) 
        {
            printf("Key shouldn't contain duplicate characters\n");
            return 1;
        }
    }
)

You can use strchr() to detect if a given character is duplicated in the rest of the string:

#include <cs50.h>
#include <ctype.h>
#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]) {
    if (argc != 2) {
        printf("Usage: ./substitution key\n");
        return 1;
    } else {
        char *key = argv[1];
        int len = strlen(key]);
        if (len != 26) {
            printf("Key must contain 26 characters.\n");
            return 1;
        } else {
            for (int i = 0; i < len; i++) {
                unsigned char c = key[i];
                if (!isalpha(c)) {
                    printf("The key must contain only letters.\n");
                    return 1;
                }
                if (strchr(key + i + 1, c) {
                    printf("The key must not contain duplicate letters.\n");
                    return 1;
                }
            }
        }
    }
    return 0;
}

Alternately, you can use an array of indicators to perform the check in a single pass. This method is preferred if you want to ignore the character case.

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

int main(int argc, char *argv[]) {
    if (argc != 2) {
        printf("Usage: ./substitution key\n");
        return 1;
    } else {
        char *key = argv[1];
        unsigned char seen['Z' + 1 - 'A'] = { 0 };
        int i;
        for (i = 0; key[i] != '\0'; i++) {
            unsigned char c = key[i];
            if (!isalpha(c)) {
                printf("The key must contain only letters.\n");
                return 1;
            }
            if (seen[toupper(c) - 'A']++ != 0) {
                printf("The key must not contain duplicate letters.\n");
                return 1;
            }
        }
        if (i != 26) {
            printf("Key must contain all 26 letters.\n");
            return 1;
        }
    }
    return 0;
}

Your can first sort the characters in the string so that any repeated letters will be right next to each other. So after sorting, scan through your sorted string with a variable holding your currently scanned character. If the next character you scan is the same as the last one that you scanned, you have a repeat character

You can consider to use strchr(). I get an example from https://www.ibm.com/support/knowledgecenter/en/ssw_ibm_i_72/rtref/strchr.htm

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

#define SIZE 40

int main(void) {
  char buffer1[SIZE] = "computer program";
  char * ptr;
  int    ch = 'p';

  ptr = strchr(buffer1, ch);
  printf("The first occurrence of %c in '%s' is '%s'\n",
          ch, buffer1, ptr );

}

/*****************  Output should be similar to:  *****************
The first occurrence of p in 'computer program' is 'puter program'
*/

There is many ways to do it, but the one I like is to first sort the string and then go through it and delete or copy the string to another string/array and u can do that by

s = "bedezdbe"

after u sort it

s = "bbddeeez"

And to check it

if(s[i-1]!=s[i])

I'm assuming you are not yet in Datastructures and Algorithms. If so I would do a binary insertion sort and when you find a repeat you just printf your error message and return a 1 like you have done for the rest of the errors.

That said, that might be over your head right now. So in that case, just go with a bruteforce method without regard for optimization.

1. Start with the first character, loop through the rest of the string to see if it matches.

2. If it matches, error message

3. if there is no match, then repeat for the next character.

The code below checks if there are any repeating characters by this brute force method mentioned. You'll need to adapt it to your code of course.

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

bool itrepeats(char c, char *restofstring, int sublen){
    for(int j=0; j<sublen; j++){
        if(c==*restofstring) return true;
        restofstring++;
    }
    return false;
}

int main (int argc, char *argv[])
{
    int len = strlen(argv[1]);
    char *substr = argv[1];
    
    //loop through all characters
    for(int i=0;i<len;i++){
        //get a character
        char c = argv[1][i];
        //start of substring after c
        substr++;
        //length of that substring
        int substrlen = len-(i+1);
        //check for c in rest of string
        if(itrepeats(c, substr, substrlen)){
            printf("%c repeats, not allowed!\n",c);
            return 1;
        }
        else printf("all good!\n");
    }
}

I've Done that before so here is my solution to this problem: Every time we found the current argv[1]'s character in the string we gonna add 1 to the counter

in normal cases where the argv[1] is normal, counter should be 26 at the end.

but if there is a repeated character, counter should be more than 26 in the end.

int counter = 0;
for (int i = 0; i < len; i++)
        {
            char argu = argv[1][i];
            // make sure that the key doesnt accept duplicated characters in key.
            for (int k = 0; k < len; k++)
            {
                if (argu == argv[1][k])
                {
                    counter++;
                }
                if (counter > 26)
                {
                    printf("Opps! no duplicated characters!\n");
                    return 1;
                }
            }
        }

#include<stdio.h>
int main(){
char  string[3] = {'c','j','j'};
for(int i = 0;i<3;i++){
    for(int j= i+1;j<3;j++){
        if(string[i] == string[j]){
            printf("position %d char %c\n",i,string[i]);
        }
    }
}

}