g0701_0800.s0790_domino_and_tromino_tiling.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g0701_0800.s0790_domino_and_tromino_tiling;

// #Medium #Dynamic_Programming #2022_03_26_Time_0_ms_(100.00%)_Space_42_MB_(14.39%)

/**
 * 790 - Domino and Tromino Tiling\.
 *
 * Medium
 *
 * You have two types of tiles: a `2 x 1` domino shape and a tromino shape. You may rotate these shapes.
 *
 * ![](https://assets.leetcode.com/uploads/2021/07/15/lc-domino.jpg)
 *
 * Given an integer n, return _the number of ways to tile an_ `2 x n` _board_. Since the answer may be very large, return it **modulo** 10⁹ + 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 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/07/15/lc-domino1.jpg)
 *
 * **Input:** n = 3
 *
 * **Output:** 5
 *
 * **Explanation:** The five different ways are show above. 
 *
 * **Example 2:**
 *
 * **Input:** n = 1
 *
 * **Output:** 1 
 *
 * **Constraints:**
 *
 * *   `1 <= n <= 1000`
**/
public class Solution {
    public int numTilings(int n) {
        if (n == 1) {
            return 1;
        } else if (n == 2) {
            return 2;
        } else if (n == 3) {
            return 5;
        } else if (n == 4) {
            return 11;
        } else if (n == 5) {
            return 24;
        }
        long[] dp = new long[n + 1];
        dp[0] = 0;
        dp[1] = 1;
        dp[2] = 2;
        dp[3] = 5;
        dp[4] = 11;
        dp[5] = 24;
        dp[6] = 53;
        for (int i = 7; i <= n; i++) {
            dp[i] = ((dp[i - 1] * 2) + dp[i - 3]) % 1000000007;
        }
        return (int) dp[n] % 1000000007;
    }
}