Count ways to reach the n’th stair

Count ways to reach the n’th stair

There are n stairs, a person standing at the bottom wants to reach the top. The person can climb either 1 stair or 2 stairs at a time. Count the number of ways, the person can reach the top.





Consider the example shown in diagram. The value of n is 3. There are 3 ways to reach the top. The diagram is taken from Easier
Fibonacci puzzles

More Examples:

Input: n = 1
Output: 1
There is only one way to climb 1 stair

Input: n = 2
Output: 2
There are two ways: (1, 1) and (2)

Input: n = 4
Output: 5
(1, 1, 1, 1), (1, 1, 2), (2, 1, 1), (1, 2, 1), (2, 2)

We strongly recommend that
you click here and practice it, before moving on to the solution.

We can easily find recursive nature in above problem. The person can reach n’th stair from either (n-1)’th stair or from (n-2)’th stair. Let the total number of ways to reach n’t stair be ‘ways(n)’. The value of ‘ways(n)’ can be written as following.

ways(n) = ways(n-1) + ways(n-2)

The above expression is actually the expression for Fibonacci
numbers, but there is one thing to notice, the value of ways(n) is equal to fibonacci(n+1).

ways(1) = fib(2) = 1

ways(2) = fib(3) = 2

ways(3) = fib(4) = 3

So we can use function for fibonacci numbers to find the value of ways(n). Following is C++ implementation of the above idea.

Run on IDE

Output:

Number of ways = 5

The time complexity of the above implementation is exponential (golden ratio raised to power n). It can be optimized to work in O(Logn) time using the previously discussed
Fibonacci function optimizations.

Generalization of the above problem

How to count number of ways if the person can climb up to m stairs for a given value m? For example if m is 4, the person can climb 1 stair or 2 stairs or 3 stairs or 4 stairs at a time.

We can write the recurrence as following.

ways(n, m) = ways(n-1, m) + ways(n-2, m) + ... ways(n-m, m)

Following is C++ implementation of above recurrence.

Run on IDE

Output:

Number of ways = 5

The time complexity of above solution is exponential. It can be optimized to O(mn) by using dynamic programming. Following is dynamic programming based solution. We build a table res[] in bottom up manner.

Run on IDE

Output:

Number of ways = 5

This article is contributed by Abhishek. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above