Python Program to 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

Python3

# A program to count the number of ways to reach n'th stair

# Recursive program to find n'th fibonacci number
def fib(n):
    if n <= 1:
        return n
    return fib(n-1) + fib(n-2)

# returns no. of ways to reach s'th stair
def countWays(s):
    return fib(s + 1)

# Driver program

s = 4
print ("Number of ways =",countWays(s))

# Contributed by Harshit Agrawal

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. 

Python3

# A program to count the number of ways to reach n'th stair

# Recursive function used by countWays
def countWaysUtil(n, m):
    if n <= 1:
        return n
    res = 0
    i = 1
    while i<= m and i<= n:
        res = res + countWaysUtil(n-i, m)
        i = i + 1
    return res

# Returns number of ways to reach s'th stair
def countWays(s, m):
    return countWaysUtil(s + 1, m)


# Driver program
s, m = 4, 2
print ("Number of ways =", countWays(s, m))

# Contributed by Harshit Agrawal

Output:

Number of ways = 5

Time complexity: O(2ⁿ)
Auxiliary Space: O(n)

Python3

# A program to count the number of ways to reach n'th stair

# Recursive function used by countWays
def countWaysUtil(n, m):
    res = [0 for x in range(n)] # Creates list res with all elements 0
    res[0], res[1] = 1, 1

    for i in range(2, n):
        j = 1
        while j<= m and j<= i:
            res[i] = res[i] + res[i-j]
            j = j + 1
    return res[n-1]

# Returns number of ways to reach s'th stair
def countWays(s, m):
    return countWaysUtil(s + 1, m)

# Driver Program
s, m = 4, 2
print ("Number of ways =", countWays(s, m))

# Contributed by Harshit Agrawal

Output:

Number of ways = 5

Time Complexity: O(m*n)

Auxiliary Space: O(n)

Please refer complete article on Count ways to reach the n’th stair for more details!