RECURSION-LEXICOGRAPHICAL ORDER one test case not running

#1

link to ques-https://hack.codingblocks.com/contests/c/457/360

#include
#include

using namespace std;

void Print(int n)
{ cout<<0<<" ";
for(int i = 1; i <= 9; i ++){
int j = 1;
while( j <= n){
for(int m = 0; m < j ; ++ m){
if(m + j * i <= n){
cout << m + j * i << " ";
}
}
j *= 10;
}
}
}

int main()
{
int n;
cin>>n;
Print(n);
return 0;
}

#2

Hey You need to write the recursive code for this.
Think recursively.Here is my code for reference -
https://ide.codingblocks.com/#/s/16028

#3

sir i am not getting the logic behind this code…could you please help

#4

Think like after printing 0 we need to print 1 and all numbers starting with one.
Then 2 and all numbers starting with 2 and so on.
So we can do this in a for loop for digits 1 to 9.We can call a print function and pass it 1 or 2 or 3 till 9.
In Print Function : print 1 and before calling print for 2,we will call print for 10*1.
Next number is 10 and then 11,12,13…19.We can print this using a for loop
But lexicographicaly 100 comes first.So 10 will call this print function again but this time multplied by 10 so that 100 and its series can be printed

#5

there is also a simple way iterate to number then covert it to string using stringstream class the store all and sort them it will automatically sorted in an lexographical order
or u can use set to store then it is automatically lexo sorted

#6 
#include <bits/stdc++.h>
using namespace std;

void print_lexiographicall(int n, int a)
{
    // cout << 0 << " ";
    if(a > n)
        return;
    cout << a << " ";
    int i =0;
    if (a==0)
        i = 1;
    for(; i<=9; i++)
        print_lexiographicall(n, a*10 + i);
}






int main()
{
	#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    print_lexiographicall(n, 0);
    
}