# Time:  O(n)
# Space: O(1)

# 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].

# Prove:
# dp[n] = dp[n-1](|) + dp[n-2](=) + 2*(dp[n-3](「」) + ... + d[0](「 = ... = 」))
#       = dp[n-1] + dp[n-2] + dp[n-3] + dp[n-3] + 2*(dp[n-4] + ... + d[0])
#       = dp[n-1] + dp[n-3] + (dp[n-2] + dp[n-3] + 2*(dp[n-4] + ... + d[0])
#       = dp[n-1] + dp[n-3] + dp[n-1]
#       = 2*dp[n-1] + dp[n-3]

class Solution(object):
    def numTilings(self, N):
        """
        :type N: int
        :rtype: int
        """
        M = int(1e9+7)
        dp = [1, 1, 2]
        for i in xrange(3, N+1):
            dp[i%3] = (2*dp[(i-1)%3]%M + dp[(i-3)%3])%M
        return dp[N%3]


Create domino-and-tromino-tiling.py