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
Solving the problem, and solving it using a dynamic programming solution are potentially two different things.
http://en.wikipedia.org/wiki/Dynamic_programming
In general, to solve a given problem, we need to solve different parts of the problem (subproblems), then combine the solutions of the subproblems to reach an overall solution. Often, many of these subproblems are really the same. The dynamic programming approach seeks to solve each subproblem only once, thus reducing the number of computations
This leads me to believe you want to look for a solution that is both Recursive, and uses the Memo Design Pattern. Recursion solves a problem by breaking it into sub-problems, and the Memo design pattern allows you to cache answers, thus avoiding re-calculation. (Note that there are probably cache implementations that aren't the Memo design pattern, and you could use one of those as well).
Solving:
The first step I would take would be to solve some set of problems by hand, with varying or increasing sizes of N. This will give you a pattern to help you figure out a solution. Start with N = 1, through N = 5. (as others have stated, it may be a form of the fibbonacci sequence, but I would determine this for myself before calling the problem solved and understood).
From there, I would try to make a generalized solution that used recursion. Recursion solves a problem by breaking it into sub-problems.
From there, I would try to make a cache of previous problem inputs to the corresponding output, hence memoizing it, and making a solution that involved "Dynamic Programming".
I.e., maybe the inputs to one of your functions are 2, 5, and the correct result was 7. Make some function that looks this up from an existing list or dictionary (based on the input). It will look for a call that was made with the inputs 2, 5. If it doesn't find it, call the function to calculate it, then store it and return the answer (7). If it does find it, don't bother calculating it, and return the previously calculated answer.