Given an integer **N**, the task is to paint a grid of size **N x 3** using colors **Red**, **Yellow**, or **Green** while making such that no pair of adjacent cells has the same* *color. Print the number of distinct ways in which it is possible

**Examples:**

Input:N = 1Output:12Explanation:

Following 12 possible ways to paint the grid exists:

- Red, Yellow, Red
- Yellow, Red, Yellow
- Green, Red, Yellow
- Red, Yellow, Green
- Yellow, Red, Green
- Green, Red, Green
- Red, Green, Red
- Yellow, Green, Red
- Green, Yellow, Red
- Red, Green, Yellow
- Yellow, Green, Yellow
- Green, Yellow, Green

Input:N = 2Output:102

**Approach:** Follow the steps below to solve the problem:

- Ways to color a row can be split into the following two categories:
- The leftmost and rightmost cells are of the same color.
- The leftmost and rightmost cells are of different colors.

- Considering the first case:
- Six possible ways exist to paint the row such that the leftmost and rightmost colors are of the same.
- For every color occupying both the leftmost and rightmost cell, there exists two different colors with which the middle row can be colored.

- Considering the second case:
- Six possible ways exist to paint the leftmost and rightmost colors are different.
- Three choices for the left cell, two choices for the middle, and fill the rightmost cell with the only remaining color. Therefore, the total number of possibilities is
**3*2*1 = 6**.

- Now, for the subsequent cells, look at the following example:
- If the previous row is painted as
**Red Green Red**, then there are the following five valid ways to color the current row:**{Green Red Green}****{Green Red Yellow}****{Green Yellow Green}****{Yellow Red Green}****{Yellow Red Yellow}**

- From the above observation, it is clear that three possibilities have ends with the same color, and two possibilities have ends with different colors.
- If the previous row is colored
**Red Green Yellow**, the following four possibilities of coloring the current row exists:**{Green Red Green}****{Green Yellow Red}****{Green Yellow Green}****{Yellow Red Green}**

- From the above observation, it is clear that possibilities have ends the same color, and two possibilities have ends with different colors.

- If the previous row is painted as
- Therefore, based on the above observations, the following recurrence relation can be defined for the number of ways to paint the
**N**rows:

Count of ways to color current row having ends of same color S

_{N+1 }= 3 * S_{N }+ 2D_{N}Count of ways to color current row having ends of different colors D

_{N+1}= 2 * S_{N }+ 2D_{N}Total number of ways to paint all N rows is equal to the sum of S

_{N}and D_{N}.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count the number` `// of ways to paint N * 3 grid` `// based on given conditions` `void` `waysToPaint(` `int` `n)` `{` ` ` `// Count of ways to pain a` ` ` `// row with same colored ends` ` ` `int` `same = 6;` ` ` `// Count of ways to pain a row` ` ` `// with different colored ends` ` ` `int` `diff = 6;` ` ` `// Traverse up to (N - 1)th row` ` ` `for` `(` `int` `i = 0; i < n - 1; i++) {` ` ` `// Calculate the count of ways` ` ` `// to paint the current row` ` ` `// For same colored ends` ` ` `same = 3 * same + 2 * diff;` ` ` `// For different colored ends` ` ` `diff = 2 * same + 2 * diff;` ` ` `}` ` ` ` ` `// Print the total number of ways` ` ` `cout << (same + diff);` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 1;` ` ` `waysToPaint(N);` `}` `// This code is contributed by ukasp.` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;` `class` `GFG {` ` ` `// Function to count the number` ` ` `// of ways to paint N * 3 grid` ` ` `// based on given conditions` ` ` `static` `void` `waysToPaint(` `int` `n)` ` ` `{` ` ` `// Count of ways to pain a` ` ` `// row with same colored ends` ` ` `long` `same = ` `6` `;` ` ` `// Count of ways to pain a row` ` ` `// with different colored ends` ` ` `long` `diff = ` `6` `;` ` ` `// Traverse up to (N - 1)th row` ` ` `for` `(` `int` `i = ` `0` `; i < n - ` `1` `; i++) {` ` ` `// Calculate the count of ways` ` ` `// to paint the current row` ` ` `// For same colored ends` ` ` `same = ` `3` `* same + ` `2` `* diff;` ` ` `// For different colored ends` ` ` `diff = ` `2` `* same + ` `2` `* diff;` ` ` `}` ` ` `// Print the total number of ways` ` ` `System.out.println(same + diff);` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `1` `;` ` ` `// Function call` ` ` `waysToPaint(N);` ` ` `}` `}` |

## Python3

`# Python3 program for the above approach` `# Function to count the number` `# of ways to paint N * 3 grid` `# based on given conditions` `def` `waysToPaint(n):` ` ` `# Count of ways to pain a` ` ` `# row with same colored ends` ` ` `same ` `=` `6` ` ` ` ` `# Count of ways to pain a row` ` ` `# with different colored ends` ` ` `diff ` `=` `6` ` ` `# Traverse up to (N - 1)th row` ` ` `for` `_ ` `in` `range` `(n ` `-` `1` `):` ` ` `# Calculate the count of ways` ` ` `# to paint the current row` ` ` ` ` `# For same colored ends` ` ` `same ` `=` `3` `*` `same ` `+` `2` `*` `diff` ` ` ` ` `# For different colored ends` ` ` `diff ` `=` `2` `*` `same ` `+` `2` `*` `diff` ` ` `# Print the total number of ways` ` ` `print` `(same ` `+` `diff)` `# Driver Code` `N ` `=` `1` `waysToPaint(N)` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to count the number` `// of ways to paint N * 3 grid` `// based on given conditions` `static` `void` `waysToPaint(` `int` `n)` `{` ` ` ` ` `// Count of ways to pain a` ` ` `// row with same colored ends` ` ` `int` `same = 6;` ` ` `// Count of ways to pain a row` ` ` `// with different colored ends` ` ` `int` `diff = 6;` ` ` `// Traverse up to (N - 1)th row` ` ` `for` `(` `int` `i = 0; i < n - 1; i++)` ` ` `{` ` ` ` ` `// Calculate the count of ways` ` ` `// to paint the current row` ` ` `// For same colored ends` ` ` `same = 3 * same + 2 * diff;` ` ` `// For different colored ends` ` ` `diff = 2 * same + 2 * diff;` ` ` `}` ` ` ` ` `// Print the total number of ways` ` ` `Console.WriteLine(same + diff);` `}` `// Driver code` `static` `public` `void` `Main()` `{` ` ` `int` `N = 1;` ` ` ` ` `waysToPaint(N);` `}` `}` `// This code is contributed by offbeat` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to count the number` `// of ways to paint N * 3 grid` `// based on given conditions` `function` `waysToPaint(n)` `{` ` ` ` ` `// Count of ways to pain a` ` ` `// row with same colored ends` ` ` `var` `same = 6;` ` ` `// Count of ways to pain a row` ` ` `// with different colored ends` ` ` `var` `diff = 6;` ` ` `// Traverse up to (N - 1)th row` ` ` `for` `(` `var` `i = 0; i < n - 1; i++)` ` ` `{` ` ` ` ` `// Calculate the count of ways` ` ` `// to paint the current row` ` ` `// For same colored ends` ` ` `same = 3 * same + 2 * diff;` ` ` `// For different colored ends` ` ` `diff = 2 * same + 2 * diff;` ` ` `}` ` ` `// Print the total number of ways` ` ` `document.write(same + diff);` `}` `// Driver code` `var` `N = 1;` `// Function call` `waysToPaint(N);` ` ` `// This code is contributed by Khushboogoyal499` `</script>` |

**Output:**

12

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

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