O(9*(n+m)*4^{m-1}): Backtracking  + dynamic programming
We have m rows and n columns; we need to determine how many colorings of the grid exist where no two 4-directionally adjacent cells have the same color.\n\nGENE
We have `m` rows and `n` columns; we need to determine how many colorings of the grid exist where no two 4-directionally adjacent cells have the same color.\n\n**GENERATING VALID COLUMNS: BACKTRACKING**\nWe first generate the set of all possible "valid columns", i.e. any column where no two consecutive rows are the same color. A column will consist of `m <= 5` cells, so we have:\n- 3 choices for the top cell\n- 2 choices for the remaining cells (any color except that of the cell on top of it)\n\nSo we are talking about `k = 3 * 2^{m-1}` possible columns to generate. This can be easily accomplished through backtracking, so we leave that to the reader. Denote this set `Valid` moving forward.\n\n**PARTITIONING SAMPLE SPACE: RECURRENCE RELATION**\nFor a matrix with precisely `i` columns of height/rows `m`, we consider the set:\n\n```\nS(i, j) = {set of matrices with i columns, where the last column corresponds to the jth valid column generated above} \n```\nwhere we enumerate the previously determined columns in `Valid` from `j=1, ... 3 * 2^{m-1} =: k`.\n\nThe set of `m-by-i` matrices where we don\'t have adjacent cells with same color is partitioned by `S(i, 1), ..., S(i, k)`. Therefore, we get our final result for an `m-by-n` matrix will be:\n```\n|S(n, 1)| + ... + |S(n, k)|\n```\nSo we can focus on the local problem of computing the size of each `S(i, j)`, and then add these together to get our final result.\n\nIf the last/`i`th column corresponds to valid column `j` (`j = 1, ..., k`), then we know that the only restriction we have is that the `i-1`st column needs to be one which did not have a row with the same color as the corresponding row in `j`; so we get:\n```\n|S(i, j)| = sum_{v in Valid: v and j do not have a common colored row} |S(i - 1, v)| \n```\nThus, we have our recurrence relation, to build to our final solution.\n\n**PRECOMPUTING**\nTo avoid computing whether a pair of columns in `Valid` share a row that has the same color, we precompute this pair-wise for everything in `Valid`, before our dynamic programming, to save time.\n\n\n**A NOTE ON TIME COMPLEXITY**\nWe notice that the time complexity of the above is `O((n + m) * 3^2 * 4^{m-1})`. Of course, the 3 and 2 come from the fact that we are only allowed 3 possible colors, and so  `|Valid| = 3 * (3 - 1)^{m-1} = 3 * 2^{m-1}`. If there were arbitrary `c` possible colors we could use, then `|Valid| = c * (c - 1)^{m-1}` possible valid columns, and our overall time complexity becomes `O((n + m) * c^2 * (c - 1)^{2(m - 1)})`.\n\n**CODE**\n```\nclass Solution:\n    def colorTheGrid(self, m: int, n: int) -> int:\n        mod = 10 ** 9 + 7\n        \n        # STEP 1: Generate all possible valid columns => O(m * 3 * 2^{m-1})\n        valid_cols = []\n        \n        def backtrack(curr_col: List[int]):\n            nonlocal valid_cols\n            # Copy of the array is O(m)\n            if len(curr_col) == m:\n                valid_cols.append(curr_col.copy())\n                return\n            for i in range(1, 3):\n                next_color = (curr_col[-1] + i) % 3\n                curr_col.append(next_color)\n                backtrack(curr_col)\n                curr_col.pop()\n            \n        for color in range(3):\n            backtrack([color])\n        \n        \n        # STEP 2: Determine which columns can be adjacent to one another => O(m * 3^2 * 4^{m-1})\n        k = len(valid_cols)\n        valid = [[False] * k for i in range(k)]\n        for i in range(k):\n            for j in range(i + 1, k):\n                v = True\n                for row in range(m):\n                    if valid_cols[i][row] == valid_cols[j][row]:\n                        v = False\n                        break\n                valid[i][j] = v\n                valid[j][i] = v\n        \n        \n        # STEP 3: Run dynamic programming => O(n * 3^2 * 4^{m-1})\n        dp = [[0] * k for i in range(n)]\n        for valid_col in range(k):\n            dp[0][valid_col] = 1\n        for i in range(1, n):\n            for curr_valid in range(k):\n                for prev_valid in range(k):\n                    if valid[prev_valid][curr_valid] == False:\n                        continue\n                    dp[i][curr_valid] = (dp[i][curr_valid] + dp[i - 1][prev_valid]) % mod\n        \n\t\t\n\t\t# STEP 4: sum over partition of all possible valid ending columns\n        res = 0\n        for i in range(k):\n            res = (res + dp[n - 1][i]) % mod\n        return res\n```\n
[]