Consider we have a digit d, and the upper limit n. we have to find all numbers that contains d in range 0 to n. So if n = 20, and digit is 3, then the numbers will be [3, 13].
To solve this problem, we will take every number as string, then if the digit is present in the string, the number will be printed, otherwise ignored.
Example
#include<iostream>
using namespace std;
int getAllNumWithDigit(int n, int d) {
string str = "";
str += to_string(d);
char ch = str[0];
string p = "";
p += ch;
for (int i = 0; i <= n; i++) {
str = "";
str = str + to_string(i);
int index = str.find(p);
if (i == d || index!=-1)
cout << (i) << " ";
}
}
int main() {
int n = 100; int d = 3;
getAllNumWithDigit(n, d);
}Output
3 13 23 30 31 32 33 34 35 36 37 38 39 43 53 63 73 83 93