There is a wall of size 4xN. You have infinite supply of bricks of size 4x1 and 1x4 in the house. How many ways are there to fill out a wall of size 4xN? Thought ProcessThis can be addressed by dynamic programming. The characteristics of optimal substructure, recursive nature, and bottomup approach were mentioned in this post. Let's say, number of ways to fill out a wall of size 4xN with infinite supply of 4x1 and 1x4 bricks is said to be f(N). Per common sense, we know: f(1) = 1 f(2) = 1 f(3) = 1 f(4) = 2 Also, when you think about it when filling out 4xN with f(N) ways of doing it, you can break f(N), when N >= 4, down to 2 substructures as below. Given the fact that one 4x1 can goes either horizontally or vertically. When it goes horizontally, there's only one way as shown in f(N1). But when it goes vertically, all other 3 bricks may need to go vertically as well as seen in f(N4) below. Code Samplepublic static int numberOfWays(int N){
1 Comment
Pushpender
10/1/2016 12:18:38 pm
Nice solution
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