How can you print 1 to 10 & 10 to 1 by using recursion with single variable

Question

I am preparing for my interview tomorrow -- I need the answer to this question:

How can you print 1 to 10 & 10 to 1 by using recursion with single variable

Answer 1

You should be able to figure this out yourself.

Hint: Make a method that takes a 10 as a parameter, then prints the parameter and calls itself with 9.

Answer 2

I'm going to get downvoted I just know it but here is (a) solution. Not the best one but you should be able to make it better yourself.

class Program
{
    static void Main(string[] args)
    {
        printNumDown(10);
    }

    static void printNumDown(int num)
    {
        Console.WriteLine(num.ToString());
        if (num > 1)
            printNumDown(num - 1);
        else
            printNumUp(num + 1);
    }

    static void printNumUp(int num)
    {
        Console.WriteLine(num.ToString());
        if (num < 10)
            printNumUp(num + 1);
    }
}

Answer 3

Here's a sneaky way:

#include <stdio.h>

static void recur_up (int n) {
    if (n > 1)
        recur_up (n - 1);
    printf ("%d\n", n);
}

static void recur_down (int n) {
    printf ("%d\n", n);
    if (n > 1)
        recur_down (n - 1);
}

int main (void) {
    recur_up (10);
    recur_down (10);
    return 0;
}

which generates:

1
2
3
4
5
6
7
8
9
10
10
9
8
7
6
5
4
3
2
1

---

It would have been a lot more elegant going down then up since you could do that with a single function:

static void recur_both (int n) {
    printf ("%d\n", n);
    if (n > 1)
        recur_down (n - 1);
    printf ("%d\n", n);
}

Answer 4

void print_recursive(int n) { 
    printf("%d\n", n);
    if (n < 10)
        print_recursive(n+1);
    printf("%d\n", n);
}

Answer 5

With one function and one variable only:

void recurs(int num) {
    printf("%d\n", num);
    if (num < 10) {
        recurs(num + 1);
    }
    printf("%d\n", num);
}

int main() {
    recurs(1);
    return 0;
}

Answer 6

Javascript version:

printNumber(1);

function printNumber(num){
  document.write(num);
  if (num < 10) 
     printNumber(num + 1);
  document.write(num);
}

Answer 7

Here's one in Ruby:

puts (r = ->n=1 { if n<=10 then [n] + r.(n+1) + [n] else [] end }).()

Note: before you conclude that Ruby is an unreadable mess even worse than Perl, let me assure you: this is not idiomatic Ruby. Idiomatic Ruby would be more like

def recursive_count_up_and_down(n=1)
  return [] unless n<=10
  [n] +
  recursive_count_up_and_down(n + 1) +
  [n]
end

puts recursive_count_up_and_down

Of course, Ruby is an imperative language, so doing it really idiomatically would not use recursion:

1.upto(10) do |i| puts i end
10.downto(1) do |i| puts i end

Here's another neat one that unfortunately doesn't use recursion, either:

puts Array.new(20) {|i| if i < 10 then i+1 else 20-i end }

BTW: all the solutions so far, including mine, are actually cheating, because technically they use two variables, because

function foo {}

is equivalent to

var foo = λ{}

So, in my example above, there are two variables: recursive_count_up_and_down and n. We could eliminate both of those by writing in a tacit point-free style (for example in the SK-calculus), but I'll leave that as an exercise to the reader. (Meaning I can't figure it out :-) )

Answer 8

Why are you guys all being so difficult? In Pseudocode:

function recurfunc(n) {
    if (n < 10) {
    echo (-1 * (floor(abs(n)) - 10));
        recurfunc(n+1);
    }
}

Then call recurfunc with -9.5 as its start.

Seems kind of obvious to me that the answer is using absolute value.