/*
VIM: let b:lcppflags="-std=c++14 -O2 -pthread -I."
VIM: let b:wcppflags="/O2 /EHsc /DWIN32"
VIM-: let b:cppflags=g:Iboost.g:Itbb
VIM-: let b:ldflags=g:Lboost.g:Ltbb
VIM-: let b:ldlibpath=g:Bboost.g:Btbb
VIM-: let b:argv=""
*/
#include <iostream>
#include <exception>
#include <chrono>
#include <vector>
#include <algorithm>

/*Given N and M find all stepping numbers in range N to M

The stepping number:

A number is called as a stepping number if the adjacent digits have a difference of 1.
e.g 123 or 101 is stepping number, but 358 is not a stepping number

Example:

N = 10, M = 20
all stepping numbers are 10 , 12 

Return the numbers in sorted order.
*/

/*
 * Solution: complexity is linear actually. O(n)
 *           So, a better solution I guess will be to interate over the range
 *           and check each number on condition.
 */


void f( std::vector<int>& v, int A, int B, int n )
{
    if ( A <= n && n <= B ) {
        v.push_back(n);
    }
    else if ( n > B ) {
        return;
    }

    int d = n % 10;
    if ( d + 1 <= 9 ){
        f( v, A, B, n*10+d+1);
    }
    if ( d - 1 >= 0 ){
        f( v, A, B, n*10+d-1);
    }
}

std::vector<int> stepnum(int A, int B)
{
    std::vector<int> v;

    if ( A <= 0 && 0 <= B ){
        v.push_back(0);
    }

    for ( int i = 1; i <= 9; ++i ){
        f( v, A, B, i );
    }

    std::sort(v.begin(), v.end());

    return v;
}

void print( const std::vector<int>& A )
{
    for (auto i: A){
        std::cout << i << " ";
    }
    std::cout << std::endl;
}

int main ( void )
{try{
    auto begin = std::chrono::high_resolution_clock::now();


    //......
    print( stepnum(10, 20) );
    print( stepnum(0, 1000) );

    auto end = std::chrono::high_resolution_clock::now();
    std::chrono::duration<double> seconds = end - begin;
    std::cout << "Time: " << seconds.count() << std::endl;
    return 0;
}
catch ( const std::exception& e )
{
    std::cerr << std::endl
            << "std::exception(\"" << e.what() << "\")." << std::endl;
    return 2;
}
catch ( ... )
{
    std::cerr  << std::endl
            << "unknown exception." << std::endl;
    return 1;
}}