How do I extract missing digits from an integer? - c++

Question

I'm trying to get the following output:
Enter integer: 87240
Missing digits: 1 3 5 6 9
Here is my code so far:
// Extract numbers inside integer n
while (numOfDigits > 0)
{
int digit = n % 10;
int missing = 0;
while ((digit != missing) && (missing < 10))
{
cout << missing << " ";
missing++;
}
numOfDigits--;
n /= 10;
}
which prints out
Enter integer: 87240
Missing digits: 0 1 2 3 0 1 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7
Is there any way to go about doing this without using an array?

Answer 1

Here is an implementation that uses a single integer (acc) to remember which digits had been seen:
#include <iostream>
using uuint = unsigned long long int;
uuint p(uuint n) { ++n; return n * n - n + 41; }
void print_missing(uuint n)
{
std::cout << "Number " << n << " is missing the following digits:";
uuint acc = 1;
while (n)
{
uuint q = p(n % 10);
if (acc % q != 0) acc *= q;
n /= 10;
}
for (int i = 0; i != 10; ++i)
{
if (acc % p(i) != 0) std::cout << " " << i;
}
std::cout << "\n";
}
int main()
{
for (uuint n; std::cin >> n; )
print_missing(n);
}

Answer 2

You can use a for loop, no need for an array:
#include <string>
#include <iostream>
#include <cstdlib>
int main()
{
std::string integer;
std::cout << "Enter integer: ";
std::getline(std::cin, integer);
for (unsigned int i = 0; i < 10; ++i)
{
const char c = '0' + i;
if (integer.find(c) == std::string::npos)
{
cout << i << " ";
}
}
return EXIT_SUCCESS;
}
The above version keeps the number in text form, easier to search for digits.
You could use a nested loop to check every digit of the number.

Answer 3

Are you interested in using a string instead of an array, perchance? Those are more or less the same idea, but given the input string n we could find any unused digits like this:
const auto digits = "0123456789"s;
sort(begin(n), end(n));
set_difference(cbegin(n), cend(n), cbegin(digits), cend(digits), ostream_iterator<char>(cout, " "));
Live Example

Answer 4

Here the version using a single integer to accumulate the results (still, technically array of bits):
#include <iostream>
#include <cmath>
void check(int n)
{
unsigned short bits = 0;
while (n) {
bits |= 1 << std::abs(n % 10);
n /= 10;
}
std::cout << "Missing digits:";
for (unsigned i = 0; i != 10; ++i) {
if (((bits >> i) & 1) == 0)
std::cout << " " << i;
}
std::cout << std::endl;
}
int main()
{
int n;
std::cout << "Enter a number: ";
std::cin >> n;
check(n);
return 0;
}

Answer 5

No Arrays in this one.
#include <stdio.h>
int NumHasDigit(int num, int digit)
{
while (num)
{
if (num%10 == digit)
{
return 1;
}
num /= 10;
}
return 0;
}
int main(void) {
int x;
printf("Enter int: ");
scanf("%d\n", &x);
printf("Missing Digits:\n");
for(int i=0; i<10; ++i)
{
if (!NumHasDigit(x,i)) printf("%d ", i);
}
return 0;
}
Sample Input:
34567
Sample Output:
Missing Digits
0 1 2 8 9