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

Method 1: The first method uses the technique of recursion to solve this problem.
Approach: We can easily find the recursive nature in the above problem. The person can reach n^th stair from either (n-1)^th stair or from (n-2)^th stair. Hence, for each stair n, we try to find out the number of ways to reach n-1^th stair and n-2^th stair and add them to give the answer for the n^th stair. Therefore the expression for such an approach comes out to be : 

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

For a better understanding, let’s refer to the recursion tree below -: 

Input: N = 4

                  fib(5)
           '3'  /        \   '2'
               /          \
           fib(4)         fib(3)
     '2'  /      \ '1'   /      \  
         /        \     /        \ 
     fib(3)     fib(2)fib(2)      fib(1) 
     /    \ '1' /   \ '0'
'1' /   '1'\   /     \ 
   /        \ fib(1) fib(0) 
fib(2)     fib(1)

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

Output: 

Number of ways = 5

Complexity Analysis: 

• Time Complexity: O(2^n) 
  The time complexity of the above implementation is exponential (golden ratio raised to power n) due to redundant calculations.It can be optimized to work in O(Logn) time using the previously discussed Fibonacci function optimizations.
• Auxiliary Space: O(1)

Generalization of the Problem 
How to count the 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.

Approach: For the generalization of above approach the following recursive relation can be used. 

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

In this approach to reach n^th stair, try climbing all possible number of stairs lesser than equal to n from present stair.

Following is the implementation of above recurrence. 

Output: 

Number of ways = 5

Complexity Analysis: 

• Time Complexity: O(2^n) 
  The time complexity of the above implementation is exponential (golden ratio raised to power n) due to redundant calculations. It can be optimized to O(m*n) by using dynamic programming.
• Auxiliary Space: O(1)

Method 2: This method uses the technique of Dynamic Programming to arrive at the solution.

Approach: We create a table res[] in bottom up manner using the following relation: 

res[i] = res[i] + res[i-j] for every (i-j) >= 0

such that the i^th index of the array will contain the number of ways required to reach the i^th step considering all the possibilities of climbing (i.e. from 1 to i).

Below code implements the above approach: 

Output: 

Number of ways = 5

Complexity Analysis: 

• Time Complexity: O(m*n)
• Auxiliary Space: O(n)

Method 3: The third method uses the technique of Sliding Window to arrive at the solution.
Approach: This method efficiently implements the above Dynamic Programming approach. 
In this approach for the i^th stair, we keep a window of sum of last m possible stairs from which we can climb to the i^th stair. Instead of running an inner loop, we maintain the result of the inner loop in a temporary variable. We remove the elements of the previous window and add the element of the current window and update the sum.

Below code implements the above idea 

Output:  

Number of ways = 13

Complexity Analysis: 
 

• Time Complexity: O(n)
• Auxiliary Space: O(n)

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

Method 4: The fourth method uses simple mathematics but this is only applicable for this problem if (Order does not matter) while counting steps.

Approach: In This method we simply count the number of sets having 2.

Output:

Number of ways when order of steps does not matter is : 3

Complexity Analysis:

• Time Complexity : O(1)
• Space Complexity  : O(1)

Note: This Method is only applicable for the question Count ways to N’th Stair(Order does not matter) .

Order does not matter means for n = 4  {1 2 1}  ,{2 1 1}  , {1 1 2} are considered same.

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