String to space separated integer - c

Question

I have a string with integers in it (in descending order) then output should be space separated integers. Integers are not negative.
INPUT: 9876543 OUTPUT: 9 8 7 6 5 4 3
INPUT: 109876543 OUTPUT: 10 9 8 7 6 5 4 3
INPUT: 400399398397 OUTPUT: 400 399 398 397
So I tried using sscanf() but was not able to get the desired result, this is the code I tried:
fgets(s1,100,stdin); // Get string
while(sscanf(data1,"%d%n",&m1,&len)==1){
b[i] = m1; // Store the integers in the array
data1 += len;
i += 1;
}
How can I achieve the desired result?

Answer 1

Continuing from the comments and the additional answer, to parse and separate the string into a space separated series of integers decreasing by one, there are probably a number of differing approaches you can take. The biggest design question is whether you start with the length of the input string, cut it in half and then work backwards decreasing the number of digits you check for adjacent values by one -- or whether you start at the beginning and work toward the end incrementing the number of digits being considered along the way.
Regardless of the direction you choose, the twist is handling/checking adjacent values with a different number of digits. Your second example, 109876543, hits at the heart of this twist, where you must code a way to check the 2-digit value 10 against the next single-digit value in the series 9. There is just no pretty way to do this. One reasonable way is to simply compute the smallest number that can be represented by n-digits (e.g. 10, 100, 1000, ...). Essentially 10^(n-1) (where we let int expn = n - 1;). If your first value v1 is equal to 10^(n-1), then reduce the number of characters you consider for the next smallest values. Something like the following:
while (expn--) /* loop to build 10 ^ (n-1) */
x10 *= 10; /* compute 10 ^ (n-1), 10, 100 */
if (v1 == x10) /* compare against v1 */
n--; /* reduce t2 by 1-char/digit */
The remainder of the task is just basically a brute force check with a minimum number of validations necessary to protect array bounds, while handling adding values to your integer array (or however you want to store values until you validate or invalidate the remaining characters in the string) while you work your way through the remaining characters.
Putting all the pieces together, and noting there are many, many ways to code this, this example being only one, you could do something similar to the following. Note, the code simply handles the conversion from ASCII to int in the single-digit series case by subtracting '0' from the character value, for multi-digit conversions, strtol is used with a validation check of errno. The code works from beginning to end of the string incrementing the number of digits checked until the end of the string is reached. If a solution is found, a space-separated list of integers is output, otherwise, "no solution found." is output. The code is commented to help you work though it.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#define MAXI 256
int main (int argc, char **argv) {
int a[MAXI] = {0}, i = 1, idx = 0, n = 1, len;
char *p = argc > 1 ? argv[1] : "9876543";
printf ("string : %s\n", p); /* original input string */
len = (int)strlen (p); /* get length */
while (i + n <= len && n < len) { /* loop until conditions met */
if (n >= MAXI) { /* protect int array bounds */
fprintf (stderr, "warning: array full, %d elements filled.\n", n);
break;
}
if (n == 1) { /* handle single digits series */
if (p[i - 1] == p[i] + 1) { /* previous equal current + 1? */
if (!idx) /* if array index == 0 */
a[idx++] = p[i - 1] - '0'; /* store first integer */
a[idx++] = p[i] - '0'; /* store current integer */
i++; /* increment string index */
}
else
n++, i = n, idx = 0; /* increment n-digits to check */
} /* set i = n, zero array index */
else { /* handle multi-digit values */
char t1[MAXI] = "", t2[MAXI] = ""; /* tmp strings for values */
int v1 = 0, v2 = 0, /* tmp for coverted values */
expn = n - 1, x10 = 1, /* 10 ^ expn for n-- test */
norig = n; /* n to restore on no match */
strncpy (t1, p + i - n, n); /* copy n-digits for 1st value */
errno = 0;
v1 = (int) strtol (t1, NULL, 10); /* convert to int/validate */
if (errno) {
fprintf (stderr, "error: failed conversion, i: %d, n: %d\n",
i, n);
return 1;
}
while (expn--) /* loop to build 10 ^ (n-1) */
x10 *= 10; /* compute 10 ^ (n-1), 10, 100 */
if (v1 == x10) /* compare against v1 */
n--; /* reduce t2 by 1-char/digit */
strncpy (t2, p + i, n); /* copy n-digits for 2nd value */
errno = 0;
v2 = (int) strtol (t2, NULL, 10); /* convert to int/validate */
if (errno) {
fprintf (stderr, "error: failed conversion, i: %d, n: %d\n",
i, n);
return 1;
}
if (v1 == v2 + 1) { /* check decreasing values */
if (!idx) /* if array index == 0 */
a[idx++] = v1; /* store first integer */
a[idx++] = v2; /* store current integer */
i += n; /* increment string index */
}
else {
n += n < norig ? 2 : 1; /* reset n if no match */
i = n; /* set string index to n */
idx = 0; /* reset array index to 0 */
}
}
}
if (idx && n < len) { /* if array has values, output */
printf ("integers :");
for (int j = 0; j < idx; j++)
printf (" %*d", n, a[j]);
putchar ('\n');
}
else
printf ("no solution found.\n");
return 0;
}
note: not all corner-cases have been evaluated and the input is presumed to contain only digits. (you are free to add the check for isdigit if you expect otherwise), further testing on your part should be done to satisfy yourself any odd-ball cases are sufficiently covered.
Example Use/Output
$ ./bin/intsepdecr
string : 9876543
integers : 9 8 7 6 5 4 3
$ ./bin/intsepdecr 109876543
string : 109876543
integers : 10 9 8 7 6 5 4 3
$ ./bin/intsepdecr 400399398397
string : 400399398397
integers : 400 399 398 397
$ ./bin/intsepdecr 400399398396
string : 400399398396
no solution found.
$ ./bin/intsepdecr 101176543
string : 101176543
no solution found.
Look things over and let me know if you have any further questions.

Answer 2

I can give a basic algorithm of how to go about this problem.
Convert the first x digits of the string into an integer (Initially x=1). You can even use a simple function like strtoi for this. (Let us say this number is N)
Find no of digits in N-1. Convert that many digits next into an integer.
Is the Converted value equal to N-1. If So, Continue and convert the rest of the string by repeating steps 2 & 3. (Need to Set N = N-1)
If not equal to N-1, Repeat from step 1, Only this time increment the number of digits converted.
Exit the program and declare a malformed string if you are not able to convert the entire string and x is greater than half the length of string.

Answer 3

Here is something i just whipped up i tested it against your cases and it works
Mind it is in c++ but it would be simple to convert to c.
string spaceNum(string in)
{
string numConvBuff;
size_t matchSize = 0;
if (in.size() == 1)
{
return to_string(in[0] - '0');
}
for (int i = 0; i < in.size() / 2; i++)
{
numConvBuff = in.substr(0, i + 1);
unsigned int numRes = stoul(numConvBuff) - 1;
string numResStr = to_string(numRes);
string n = in.substr(i + 1, numResStr.length());
if(numRes == stoul(n))
matchSize = i+1;
}
if (matchSize)
{
string out = in.substr(0, matchSize);
unsigned int numRes = stoul(out);
for (size_t i = matchSize; i < in.length();)
{
numRes--;
string numResStr = to_string(numRes);
string n = in.substr(i, numResStr.length());
out += " " + n;
i += numResStr.length();
}
return out;
}
return "";
}