# Ways to paint stairs with two colors such that two adjacent are not yellow

Given n stairs and we have 2 colour yellow and green the task is that we have to paint given stairs by given colour with condition is that we cannot paints two yellow steps directly after each other.

Examples :

```Input : n = 1
Output : 2
A single stair can be colored either
as green or yellow.

Input : n = 3
Output : 5
``` Case 1: When we have 1 stair, we can paint either yellow or green.
Case 2: When we have 2 stairs, we can paint first stair by either yellow or green but for next stair we can only paint by green because we cannot paint two yellow steps directly after each other. So total cases are three YG, GG, GY.
Case 3: When we have 3 stairs then we can paint it by in 5 ways.

If we take a closer look, we can notice that it follows Fibonacci Series. ## C++

 `// C++ Program to find the number of ways to paint stairs ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the number of ways ` `int` `ways(``int` `n) ` `{ ` `    ``int` `W[n + 1]; ` ` `  `    ``// take base case for 1 and 2 ` `    ``W = 2; ` `    ``W = 3; ` ` `  `    ``for` `(``int` `i = 3; i <= n; i++)  ` `        ``W[i] = W[i - 1] + W[i - 2]; ` `     `  `    ``return` `W[n]; ` `} ` ` `  `// Driven code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``printf``(``"%d"``, ways(n)); ` `    ``return` `0; ` `} `

## Java

 `// java Program to find the number of ` `// ways to paint stairs ` `import` `java.io.*; ` ` `  `public` `class` `GFG { ` `     `  `    ``// Function to find the number of ways ` `    ``static` `int` `ways(``int` `n) ` `    ``{ ` `        ``int` `[]W = ``new` `int``[n+``1``]; ` `     `  `        ``// take base case for 1 and 2 ` `        ``W[``1``] = ``2``; ` `        ``W[``2``] = ``3``; ` `     `  `        ``for` `(``int` `i = ``3``; i <= n; i++)  ` `            ``W[i] = W[i - ``1``] + W[i - ``2``]; ` `         `  `        ``return` `W[n]; ` `    ``} ` `     `  `    ``// Driven code ` `    ``static` `public` `void` `main (String[] args) ` `    ``{ ` `        ``int` `n = ``3``; ` `         `  `        ``System.out.println(ways(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Python3 code to find the number ` `# of ways to paint stairs ` ` `  `# Function to find the number of ways ` `def` `ways( n ): ` `    ``W ``=` `list``() ` `     `  `    ``# take base case for 1 and 2 ` `    ``W.append(``0``) ` `    ``W.append(``2``) ` `    ``W.append(``3``) ` `     `  `    ``i ``=` `3` `    ``while` `i <``=` `n: ` `        ``W.append(W[i ``-` `1``] ``+` `W[i ``-` `2``]) ` `        ``i ``=` `i ``+` `1` `         `  `    ``return` `W[n] ` ` `  `# Driver code ` `n ``=` `3` `print``(ways(n)) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

## C#

 `// C# Program to find the number of ` `// ways to paint stairs ` `using` `System; ` ` `  `public` `class` `GFG { ` `     `  `    ``// Function to find the number of ways ` `    ``static` `int` `ways(``int` `n) ` `    ``{ ` `        ``int` `[]W =``new` `int``[n+1]; ` `     `  `        ``// take base case for 1 and 2 ` `        ``W = 2; ` `        ``W = 3; ` `     `  `        ``for` `(``int` `i = 3; i <= n; i++)  ` `            ``W[i] = W[i - 1] + W[i - 2]; ` `         `  `        ``return` `W[n]; ` `    ``} ` `     `  `    ``// Driven code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `n = 3; ` `     `  `        ``Console.WriteLine(ways(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output :

```5
```

Time Complexity : O(n)
Extra Space : O(n)

We can solve this problem in O(Log n) time also using matrix exponentiation solution for n-th Fibonacci Number.

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