Counting ways to climb n steps with 1, 2, or 3 steps taken

Question

In a book I encountered following question:

Given N step stair, in how many number of ways can you climb if you use either 1, 2 or 3 steps at a time?

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Answer 1

As Geobits said there is four ways to climb 3 stairs

(1,1,1),(1,2),(2,1),(3)

As for why it returns 1 when n==0, this is because whenever n==0 that means that you have found exactly 1 way to climb the stairs.

Lastly this algorithm will take a very long time if you put any high number in, if you are looking for a dynamic programming approach you should create a n sized array and initialize the first three entries as 1,2,4 and then iterate through the array starting and index 3, each time set the index to array[i-1] + array[i-2] + array[i-3]. This will do the minimal amount of computation but will use a huge amount of memory.

There is a better way where you only use a 1x3 sized array, but I will leave how to do that as an exercise for the user.

Answer 2

I do not understand why 1 is being returned for n=0. If there are 0 steps then obviously we do not have to climb any and 0 should be returned.

When there are no steps you just go through without climbing, which is the one and only one way. As is pointed out in one of the comments, it can be represented as ().

For n=3 function returns 4 but i can see only 3 cases i.e. (1,1,1), (1,2), (3).

There are actually 4 cases: (1,1,1), (1,2), (2,1), (3).

Answer 3

I do not understand why 1 is being returned for n=0. If there are 0 steps then obviously we do not have to climb any and 0 should be returned.

To complement the answer by Terry, the general answer to the problem is the tribonacci(n+2) sequence. Accordingly, for n=0, i.e. tribonacci(2), the value is 1. This is just a computational hack for the stairs problem, one that works. For a more thorough investigation, please see this answer.

You can certainly choose to reject f(n=0) = 1. If you do, you can then just use the following base case values which you will be more comfortable with. All n>3 will be recursed to these base cases.

• f(n=1) = 1
• f(n=2) = 2
• f(n=3) = 4