790. Domino and Tromino Tiling

Question

We have two types of tiles: a 2x1 domino shape, and an “L” tromino shape. These shapes may be rotated.

XX <- domino

XX <- “L” tromino X Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7.

(In a tiling, every square must be covered by a tile. Two tilings are different if and only if there are two 4-directionally adjacent cells on the board such that exactly one of the tilings has both squares occupied by a tile.)

Example:
Input: 3
Output: 5
Explanation:
The five different ways are listed below, different letters indicates different tiles:
XYZ XXZ XYY XXY XYY
XYZ YYZ XZZ XYY XXY

Note:

N will be in range [1, 1000].

Solution:

Method 1: dp

Let’s go to a simpler question, how many ways to implement 2 * N blocks only using domino?
- 2 * 1
```
1
1
```
- 2 * 2
```
11        12
22        12
```
- 2 * i: dp[i] = dp[i - 1] + dp[i - 2]
```
dp[i-1]1    +     dp[i-2]１１
dp[i-1]1    +     dp[i-2]２２
```

Come back to this question, we use dp[i][0-2]

dp[i][0]: Fill up to col i and both rows fills.
dp[i][1]: Fill up to col i and upper row fills.
dp[i][2]: Fill up to col i and bottom row fills.

State transfer function

dp[i][0] = dp[i - 1][0] + dp[i - 2][0] + dp[i - 1][1] + dp[i - 2][2]
dp[i][1] = dp[i - 2][0] + dp[i - 1][2]

dp[i][2] = dp[i - 2][0] + dp[i - 1][1]

class Solution {
public int numTilings(int N) {
    if(N == 1) return 1;
    if(N == 2) return 2;
    long[][] dp = new long[N + 1][3];
    dp[1][0] = 1; dp[2][0] = 2;
    dp[2][1] = 1;
    dp[2][2] = 1;
    int kmod = (int)1e9 + 7;
    for(int i = 3; i <= N; i++){
        dp[i][0] = (dp[i - 1][0] + dp[i - 2][0] + dp[i - 1][1] + dp[i - 1][2]) % kmod;
        dp[i][1] = (dp[i - 2][0] + dp[i - 1][2]) % kmod;
        dp[i][2] = (dp[i - 2][0] + dp[i - 1][1]) % kmod;
    }
    return (int)dp[N][0];
}
}

Reference:

花花酱 LeetCode 790. Domino and Tromino Tiling - 刷题找工作 EP171