Question

I was given the following problem in an interview:

Given a staircase with N steps, you can go up with 1 or 2 steps each time. Output all possible way you go from bottom to top.

For example:

N = 3

Output :
1 1 1
1 2
2 1

When interviewing, I just said to use dynamic programming.

S(n) = S(n-1) +1 or S(n) = S(n-1) +2

However, during the interview, I didn’t write very good code for this. How would you code up a solution to this problem?

Thanks indeed!

Answer 1

You can generalize your recursive function to also take already made moves.

void steps(n, alreadyTakenSteps) {
    if (n == 0) {
        print already taken steps
    }
    if (n >= 1) {
        steps(n - 1, alreadyTakenSteps.append(1));
    }
    if (n >= 2) {
        steps(n - 2, alreadyTakenSteps.append(2));
    }
}

It’s not really the code, more of a pseudocode, but it should give you an idea.