Given a 3 X n board, find the number of ways to color it using at most 4 colors such that no two adjacent boxes have the same color. Diagonal neighbors are not treated as adjacent boxes.

Output the ways%1000000007 as the answer grows quickly.

Constraints:

1<= n < 100000**Examples :**

Input : 1 Output : 36 We can use either a combination of 3 colors or 2 colors. Now, choosing 3 colors out of 4 is and arranging them in 3! ways, similarly choosing 2 colors out of 4 is and while arranging we can only choose which of them could be at centre, that would be 2 ways. Answer = *3! + *2! = 36 Input : 2 Output : 588

We are going to solve this using dynamic approach because when a new column is added to the board, the ways in which colors are going to be filled depends just upon the color pattern in the current column. We can only have a combination of two colors and three colors in a column. All possible new columns that can be generated is given in the image. Please consider A, B, C and D as 4 colors.

From now, we will refer 3 colors combination for a Nth column of the 3*N board as W(n) and two colors as Y(n).

We can see that each W can generate 5Y and 11W, and each Y can generate 7Y and 10W. We get two equation from here

We have two equations now,

W(n+1) = 10*Y(n)+11*W(n); Y(n+1) = 7*Y(n)+5*W(n);

## C++

`// C++ program to find number of ways` `// to color a 3 x n grid using 4 colors` `// such that no two adjacent have same` `// color` `#include <iostream>` `using` `namespace` `std;` `int` `solve(` `int` `A)` `{` ` ` ` ` `// When we to fill single column` ` ` `long` `int` `color3 = 24;` ` ` `long` `int` `color2 = 12;` ` ` `long` `int` `temp = 0;` ` ` ` ` `for` `(` `int` `i = 2; i <= A; i++)` ` ` `{` ` ` `temp = color3;` ` ` `color3 = (11 * color3 + 10 *` ` ` `color2 ) % 1000000007;` ` ` ` ` `color2 = ( 5 * temp + 7 *` ` ` `color2 ) % 1000000007;` ` ` `}` ` ` ` ` `long` `num = (color3 + color2)` ` ` `% 1000000007;` ` ` ` ` `return` `(` `int` `)num;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `num1 = 1;` ` ` `cout << solve(num1) << endl;` ` ` `int` `num2 = 2;` ` ` `cout << solve(num2) << endl;` ` ` ` ` `int` `num3 = 500;` ` ` `cout << solve(num3) << endl;` ` ` `int` `num4 = 10000;` ` ` `cout << solve(num4);` ` ` ` ` `return` `0;` `}` `// This code is contributed by vt_m.` |

## Java

`// Java program to find number of ways to color` `// a 3 x n grid using 4 colors such that no two` `// adjacent have same color.` `public` `class` `Solution {` ` ` `public` `static` `int` `solve(` `int` `A) {` ` ` `long` `color3 = ` `24` `; ` `// When we to fill single column` ` ` `long` `color2 = ` `12` `;` ` ` `long` `temp = ` `0` `;` ` ` `for` `(` `int` `i = ` `2` `; i <= A; i++) ` ` ` `{` ` ` `long` `temp = color3;` ` ` `color3 = (` `11` `* color3 + ` `10` `* color2 ) % ` `1000000007` `;` ` ` `color2 = ( ` `5` `* temp + ` `7` `* color2 ) % ` `1000000007` `;` ` ` `}` ` ` `long` `num = (color3 + color2) % ` `1000000007` `;` ` ` `return` `(` `int` `)num;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `num1 = ` `1` `;` ` ` `System.out.println(solve(num1));` ` ` `int` `num2 = ` `2` `;` ` ` `System.out.println(solve(num2));` ` ` `int` `num3 = ` `500` `;` ` ` `System.out.println(solve(num3));` ` ` `int` `num4 = ` `10000` `;` ` ` `System.out.println(solve(num4));` ` ` `}` `}` |

## Python3

`# Python 3 program to find number of ways` `# to color a 3 x n grid using 4 colors` `# such that no two adjacent have same` `# color` `def` `solve(A):` ` ` ` ` `# When we to fill single column` ` ` `color3 ` `=` `24` ` ` `color2 ` `=` `12` ` ` `temp ` `=` `0` ` ` ` ` `for` `i ` `in` `range` `(` `2` `, A ` `+` `1` `, ` `1` `):` ` ` `temp ` `=` `color3` ` ` `color3 ` `=` `(` `11` `*` `color3 ` `+` `10` `*` `color2 ) ` `%` `1000000007` ` ` ` ` `color2 ` `=` `( ` `5` `*` `temp ` `+` `7` `*` `color2 ) ` `%` `1000000007` ` ` ` ` `num ` `=` `(color3 ` `+` `color2) ` `%` `1000000007` ` ` ` ` `return` `num` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `num1 ` `=` `1` ` ` `print` `(solve(num1))` ` ` ` ` `num2 ` `=` `2` ` ` `print` `(solve(num2))` ` ` ` ` `num3 ` `=` `500` ` ` `print` `(solve(num3))` ` ` `num4 ` `=` `10000` ` ` `print` `(solve(num4))` ` ` `# This code is contributed by` `# Shashank_Sharma` |

## C#

`// C# program to find number of ways` `// to color a 3 x n grid using 4` `// colors such that no two adjacent` `// have same color.` `using` `System;` `public` `class` `GFG {` ` ` ` ` `public` `static` `int` `solve(` `int` `A)` ` ` `{` ` ` ` ` `// When we to fill single column` ` ` `long` `color3 = 24;` ` ` `long` `color2 = 12;` ` ` `long` `temp = 0;` ` ` ` ` `for` `(` `int` `i = 2; i <= A; i++)` ` ` `{` ` ` `temp = color3;` ` ` `color3 = (11 * color3 + 10` ` ` `* color2 ) % 1000000007;` ` ` ` ` `color2 = ( 5 * temp + 7` ` ` `* color2 ) % 1000000007;` ` ` `}` ` ` `long` `num = (color3 + color2)` ` ` `% 1000000007;` ` ` `return` `(` `int` `)num;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `num1 = 1;` ` ` `Console.WriteLine(solve(num1));` ` ` `int` `num2 = 2;` ` ` `Console.WriteLine(solve(num2));` ` ` `int` `num3 = 500;` ` ` `Console.WriteLine(solve(num3));` ` ` `int` `num4 = 10000;` ` ` `Console.WriteLine(solve(num4));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to find number of ways` `// to color a 3 x n grid using 4 colors` `// such that no two adjacent have same` `// color` `function` `solve(` `$A` `)` `{` ` ` ` ` `// When we to fill single column` ` ` `$color3` `= 24;` ` ` `$color2` `= 12;` ` ` `$temp` `= 0;` ` ` ` ` `for` `(` `$i` `= 2; ` `$i` `<= ` `$A` `; ` `$i` `++)` ` ` `{` ` ` `$temp` `= ` `$color3` `;` ` ` `$color3` `= (11 * ` `$color3` `+` ` ` `10 * ` `$color2` `) %` ` ` `1000000007;` ` ` ` ` `$color2` `= ( 5 * ` `$temp` `+` ` ` `7 * ` `$color2` `) %` ` ` `1000000007;` ` ` `}` ` ` ` ` `$num` `= (` `$color3` `+ ` `$color2` `) %` ` ` `1000000007;` ` ` ` ` `return` `(int)` `$num` `;` `}` `// Driver code` `$num1` `= 1;` `echo` `solve(` `$num1` `) ,` `"\n"` `;` `$num2` `= 2;` `echo` `solve(` `$num2` `) ,` `"\n"` `;` `$num3` `= 500;` `echo` `solve(` `$num3` `),` `"\n"` `;` `$num4` `= 10000;` `echo` `solve(` `$num4` `);` `// This code is contributed by m_kit.` `?>` |

## Javascript

`<script>` `// JavaScript program to find number of ways to color` `// a 3 x n grid using 4 colors such that no two` `// adjacent have same color.` ` ` `function` `solve(A) {` ` ` `let color3 = 24; ` `// When we to fill single column` ` ` `let color2 = 12;` ` ` `let temp = 0;` ` ` `for` `(let i = 2; i <= A; i++) ` ` ` `{` ` ` `let temp = color3;` ` ` `color3 = (11 * color3 + 10 * color2 ) % 1000000007;` ` ` `color2 = ( 5 * temp + 7 * color2 ) % 1000000007;` ` ` `}` ` ` `let num = (color3 + color2) % 1000000007;` ` ` `return` `num;` ` ` `}` ` ` `// Driver Code` ` ` `let num1 = 1;` ` ` `document.write(solve(num1) + ` `"<br/>"` `);` ` ` ` ` `let num2 = 2;` ` ` `document.write(solve(num2) + ` `"<br/>"` `);` ` ` ` ` `let num3 = 500;` ` ` `document.write(solve(num3) + ` `"<br/>"` `);` ` ` ` ` `let num4 = 10000;` ` ` `document.write(solve(num4));` ` ` `</script>` |

**Output :**

36 588 178599516 540460643

This article is contributed by **Panshul Garg**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend live classes with industry experts, please refer **DSA Live Classes**