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.
1<= n < 100000
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);
36 588 178599516 540460643
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