# 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 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.

## C

 `// A C program to count number of ways to reach n't stair when ` `// a person can climb 1, 2, ..m stairs at a time. ` `#include ` ` `  `// A simple recursive program to find n'th fibonacci number ` `int` `fib(``int` `n) ` `{ ` `   ``if` `(n <= 1) ` `      ``return` `n; ` `   ``return` `fib(n-1) + fib(n-2); ` `} ` ` `  `// Returns number of ways to reach s'th stair ` `int` `countWays(``int` `s) ` `{ ` `    ``return` `fib(s + 1); ` `} ` ` `  `// Driver program to test above functions ` `int` `main () ` `{ ` `  ``int` `s = 4; ` `  ``printf``(``"Number of ways = %d"``, countWays(s)); ` `  ``getchar``(); ` `  ``return` `0; ` `}`

## Java

 `class` `stairs ` `{ ` `    ``// A simple recursive program to find n'th fibonacci number ` `    ``static` `int` `fib(``int` `n) ` `    ``{ ` `       ``if` `(n <= ``1``) ` `          ``return` `n; ` `       ``return` `fib(n-``1``) + fib(n-``2``); ` `    ``} ` `     `  `    ``// Returns number of ways to reach s'th stair ` `    ``static` `int` `countWays(``int` `s) ` `    ``{ ` `        ``return` `fib(s + ``1``); ` `    ``} ` ` `  ` `  `    ``/* Driver program to test above function */`  `    ``public` `static` `void` `main (String args[]) ` `    ``{ ` `          ``int` `s = ``4``; ` `            ``System.out.println(``"Number of ways = "``+ countWays(s)); ` `    ``} ` `}``/* This code is contributed by Rajat Mishra */`

## Python

 `# A program to count the number of ways to reach n'th stair ` ` `  `# Recurssive 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 = "``, ` `print` `countWays(s) ` ` `  `# Contributed by Harshit Agrawal `

## C#

 `// C# program to count the  ` `// number of ways to reach  ` `// n'th stair  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// A simple recursive ` `    ``// program to find n'th ` `    ``// fibonacci number ` `    ``static` `int` `fib(``int` `n) ` `    ``{ ` `    ``if` `(n <= 1) ` `        ``return` `n; ` `    ``return` `fib(n - 1) +  ` `           ``fib(n - 2); ` `    ``} ` `     `  `    ``// Returns number of ways ` `    ``// to reach s'th stair ` `    ``static` `int` `countWays(``int` `s) ` `    ``{ ` `        ``return` `fib(s + 1); ` `    ``} ` ` `  `        ``// Driver Code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `s = 4; ` `        ``Console.WriteLine(``"Number of ways = "` `+  ` `                                 ``countWays(s)); ` `    ``} ` `} ` ` `  `// This code is contributed  ` `// by akt_mit `

## PHP

 ` `

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 the implementation of above recurrence.

## C

 `// A C program to count number of ways to reach n't stair when ` `// a person can climb either 1 or 2 stairs at a time ` `#include ` ` `  `// A recursive function used by countWays ` `int` `countWaysUtil(``int` `n, ``int` `m) ` `{ ` `    ``if` `(n <= 1) ` `        ``return` `n; ` `    ``int` `res = 0; ` `    ``for` `(``int` `i = 1; i<=m && i<=n; i++) ` `        ``res += countWaysUtil(n-i, m); ` `    ``return` `res; ` `} ` ` `  `// Returns number of ways to reach s'th stair ` `int` `countWays(``int` `s, ``int` `m) ` `{ ` `    ``return` `countWaysUtil(s+1, m); ` `} ` ` `  `// Driver program to test above functions- ` `int` `main () ` `{ ` `    ``int` `s = 4, m = 2; ` `    ``printf``(``"Nuber of ways = %d"``, countWays(s, m)); ` `    ``return` `0; ` `} `

## Java

 `class` `stairs ` `{ ` `    ``// A recursive function used by countWays ` `    ``static` `int` `countWaysUtil(``int` `n, ``int` `m) ` `    ``{ ` `        ``if` `(n <= ``1``) ` `            ``return` `n; ` `        ``int` `res = ``0``; ` `        ``for` `(``int` `i = ``1``; i<=m && i<=n; i++) ` `            ``res += countWaysUtil(n-i, m); ` `        ``return` `res; ` `    ``} ` `  `  `    ``// Returns number of ways to reach s'th stair ` `    ``static` `int` `countWays(``int` `s, ``int` `m) ` `    ``{ ` `        ``return` `countWaysUtil(s+``1``, m); ` `    ``} ` ` `  ` `  `    ``/* Driver program to test above function */`  `    ``public` `static` `void` `main (String args[]) ` `    ``{ ` `          ``int` `s = ``4``,m = ``2``; ` `            ``System.out.println(``"Number of ways = "``+ countWays(s,m)); ` `    ``} ` `}``/* This code is contributed by Rajat Mishra */`

## Python

 `# 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 `

## C#

 `// C# program to Count ways to reach ` `// the n’th stair ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// A recursive function used by  ` `    ``// countWays ` `    ``static` `int` `countWaysUtil(``int` `n, ``int` `m) ` `    ``{ ` `        ``if` `(n <= 1) ` `            ``return` `n; ` `        ``int` `res = 0; ` `         `  `        ``for` `(``int` `i = 1; i <= m && i <= n; i++) ` `            ``res += countWaysUtil(n-i, m); ` `        ``return` `res; ` `    ``} ` ` `  `    ``// Returns number of ways to reach ` `    ``// s'th stair ` `    ``static` `int` `countWays(``int` `s, ``int` `m) ` `    ``{ ` `        ``return` `countWaysUtil(s+1, m); ` `    ``} ` ` `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `s = 4,m = 2; ` `        ``Console.Write(``"Number of ways = "` `                           ``+ countWays(s, m)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` `

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.

## C

 `// A C program to count number of ways to reach n't stair when ` `// a person can climb 1, 2, ..m stairs at a time ` `#include ` ` `  `// A recursive function used by countWays ` `int` `countWaysUtil(``int` `n, ``int` `m) ` `{ ` `    ``int` `res[n]; ` `    ``res = 1; res = 1; ` `    ``for` `(``int` `i=2; i

## Java

 `// Java program to count number of ways to reach n't stair when ` `// a person can climb 1, 2, ..m stairs at a time ` ` `  `class` `GFG ` `{ ` `    ``// A recursive function used by countWays ` `    ``static` `int` `countWaysUtil(``int` `n, ``int` `m) ` `    ``{ ` `        ``int` `res[] = ``new` `int``[n]; ` `        ``res[``0``] = ``1``; res[``1``] = ``1``; ` `        ``for` `(``int` `i=``2``; i

## Python

 `# 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 witth 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` `"Nmber of ways ="``,countWays(s,m) ` `     `  `# Contributed by Harshit Agrawal `

## C#

 `// C# program to count number  ` `// of ways to reach n't stair when ` `// a person can climb 1, 2, ..m  ` `// stairs at a time ` `using` `System; ` `class` `GFG { ` `     `  `    ``// A recursive function ` `    ``// used by countWays ` `    ``static` `int` `countWaysUtil(``int` `n, ``int` `m) ` `    ``{ ` `        ``int` `[]res = ``new` `int``[n]; ` `        ``res = 1; res = 1; ` `        ``for` `(``int` `i = 2; i < n; i++) ` `        ``{ ` `            ``res[i] = 0; ` `            ``for` `(``int` `j = 1; j <= m && j <= i; j++) ` `                ``res[i] += res[i - j]; ` `        ``} ` `        ``return` `res[n - 1]; ` `    ``} ` `     `  `    ``// Returns number of ways  ` `    ``// to reach s'th stair ` `    ``static` `int` `countWays(``int` `s, ``int` `m) ` `    ``{ ` `        ``return` `countWaysUtil(s + 1, m); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `s = 4, m = 2; ` `        ``Console.WriteLine(``"Number of ways = "` `+ countWays(s, m)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## PHP

 ` `

Output:

`Number of ways = 5`