Generate all possible combinations of 3 digits without repetition

Here is a fairly compact snippet of code for giving the smallest unique three-digit number.

#include <stdio.h>

int main()
{
    char digit[4];

    for (int i = 0; i < 8; i++)
    {
        for (int j = i + 1; j < 9; j++)
        {
            for (int k = j + 1; k < 10; k++)
            {
                digit[0] = i + '0';
                digit[1] = j + '0';
                digit[2] = k + '0';
                digit[3] = '\0';
                printf("%s\n", digit);
            }
        }
    }

    return 0;
}

You can give that a try.

You can do:

#include <stdio.h>

int main (void) {
    char h = '0', t = '1', u = '2';

    while ((h <= '7') || (t <= '8') || (u <= '9')) {
        printf ("%c%c%c, ", h, t, u);
        u != '9' ? ++u : (t != '8' ? (++t, u = t + 1) : (++h, t = h + 1, u = t + 1));
    }
    
    return 0;
}

Output:

012, 013, 014, 015, 016, 017, 018, 019, 023, 024, 025, 026, 027, 028, 029, 034,
035, 036, 037, 038, 039, 045, 046, 047, 048, 049, 056, 057, 058, 059, 067, 068,
069, 078, 079, 089, 123, 124, 125, 126, 127, 128, 129, 134, 135, 136, 137, 138,
139, 145, 146, 147, 148, 149, 156, 157, 158, 159, 167, 168, 169, 178, 179, 189,
234, 235, 236, 237, 238, 239, 245, 246, 247, 248, 249, 256, 257, 258, 259, 267,
268, 269, 278, 279, 289, 345, 346, 347, 348, 349, 356, 357, 358, 359, 367, 368,
369, 378, 379, 389, 456, 457, 458, 459, 467, 468, 469, 478, 479, 489, 567, 568,
569, 578, 579, 589, 678, 679, 689, 789,

Explanation:

• In the program - h, t and u represents digit at hundredth location, digit at tenth location and digit at unit location respectively in a three digit number.

• Initialised h with '0', t with '1' and u with '2' because 012 is the smallest three digit number without any of the digits being repeated.

• while loop condition ((h <= '7') || (t <= '8') || (u <= '9')) because 789 is biggest three digit number which follow constraint of - smallest combination of three digits without any of the digits being repeated. Any other combination of 7, 8 and 9 will be bigger than 789 and any three digit number bigger than 789 will violat one of the given constraints.

• If you find it difficult to understand the expression with ternary operation

        u != '9' ? ++u : (t != '8' ? (++t, u = t + 1) : (++h, t = h + 1, u = t + 1));
  

  then here is simplified version using if else :

        if (u != '9') {
            ++u;
        } else {
            if (t != '8') {
                ++t;
                u = t + 1;   // due to given constraint, u will always greater than t
            } else {
                ++h;
                t = h + 1;   // due to given constraint, t will always greater than h
                u = t + 1;   // due to given constraint, u will always greater than t
            }
        }
  

Here's some annotated code that might improve the outcome.

char *sep = ""; // the separator to use. Put this line ahead of the loop

and

// meaningful names mean no comment required
// EDIT: thank you, @Chux
int unit = i      % 10;
int tens = i / 10 % 10;
int hund = i /100 % 10;

if ( unit < tens && tens <  hund )
    printf( "%s%c%c%c", sep, hund + '0', tens + '0', unit + '0' ), sep = ", ";

Notice that 'j', 'k' and 'l' have no meaning to this problem. Easy to putchar() the numbers in reverse order without noticing...

Of course, if this is not a homework assignment (Prof wouldn't believe you wrote this)...

#include <stdio.h>

int main() {
    char *sep = "", *dgts = "0123456789";
    for( int h = 0; dgts[h]; h++ )
        for( int t = h+1; dgts[t]; t++ )
            for( int u = t+1; dgts[u]; u++ )
                printf( "%s%c%c%c", sep, dgts[h], dgts[t], dgts[u] ), sep = ", ";
    putchar( '\n' );
    return 0;
}

And, to use the equivalence of array addressing and pointers.

#include <stdio.h>
int main() {
    char *h, *t, *u, *sep = "";
    for( h = "0123456789"; *h; h++ )
        for( t = h+1; *t; t++ )
            for( u = t+1; *u; u++ )
                printf( "%s%c%c%c", sep, *h, *t, *u), sep = ", ";
    putchar( '\n' );
    return 0;
}

Inspired by comment by @Chux, here is the same thing expanded to printing 4 digits. Comparison will show the few trivial alterations needed.

#include <stdio.h>
int main() {
    char *m, *h, *t, *u, *sep = "";
    for( m = "0123456789"; *m; m++ )
        for( h = m+1; *h; h++ )
            for( t = h+1; *t; t++ )
                for( u = t+1; *u; u++ )
                    printf( "%s%c%c%c%c", sep, *m, *h, *t, *u), sep = ", ";
    putchar( '\n' );
    return 0;
}

Here is how I went about my solution:

int main(void)
{
        int a, b, c;

        for (a = 0; a < 8; a++)
        {
                for (b = a + 1; b < 9; b++)
                {
                        for (c = b + 1; c < 10; c++)
                        {
                                if (a != b && a != c && b != c)
                                {
                                        putchar(a + '0');
                                        putchar(b + '0');
                                        putchar(c + '0');
                                        if (a + b + c < 24)
                                        {
                                                putchar(',');
                                                putchar(' ');
                                        }
                                }
                        }
                }
        }
        putchar('\n');
        return (0);
}

The variables in order of a, b, & c; a for 1st digit, b for 2nd and c for 3rd.

The use of b = a + 1 & c = b + 1 was an idea given to me by user @chux - Reinstate Monica in another post that linked me to this one that involved two integers instead of 3 (For two digits, the code only needs to be slightly tweaked to cater for 2 integers).

Moving on.

The if statements:

if (a != b && a != c && b != c)

and

if (a + b + c < 24)

will filter out the duplicates and ensure that the comma and space are not applied to the last combo of integers respectively (i.e. 789, which sum up to give 24).