Possible different combinations of two digits in C language [duplicate] - c

Question

I'm trying to write a program that prints all possible combinations of three digits.
And there is the constraint that:
The three digits must be different
012, 120, 102, 021, 201, and 210 are considered the same combination of the three digits 0, 1 and 2
Print only the smallest combination of three digits
The expected output is as follows
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
I'm trying to implement in C but the algorithm or any language implementation will suffice.
Here is what I've done so far but it's inacurrate.
#include <stdio.h>
/**
* main - entry point
*
* Description: display triple digits and ,
*
* Return: Always 0 (successful)
*/
int main(void)
{
int i, j, k, l;
i = 0;
while (i < 1000)
{
j = i / 100; /* hundreds */
k = (i / 10) % 10; /* tens */
l = i % 100; /* units */
if (j < k && k < l)
{
putchar(l + '0');
putchar(k + '0');
putchar(j + '0');
if (i < 789)
{
putchar(',');
putchar(' ');
}
}
i++;
}
putchar('\n');
return (0);
}

Answer 1

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.

Answer 2

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
}
}

Answer 3

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;
}

Answer 4

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).