Python Solution
\n\n# Approach\nHere\'s a step-by-step explanation of the solution:\n\n1. Generate all valid rows:\nThe generate_valid_rows function generates all valid rows of
![Screenshot 2023-03-15 at 09.12.21.png](https://assets.leetcode.com/users/images/f75a5e94-b329-4550-a02d-d5a0bd78db12_1678864659.9028294.png)\n\n# Approach\nHere\'s a step-by-step explanation of the solution:\n\n1. Generate all valid rows:\nThe generate_valid_rows function generates all valid rows of length m, with no two adjacent cells having the same color. There are 3 ** m possible row combinations, and we iterate through all of them to find the valid ones. To check the validity of a row, we ensure that no two adjacent cells have the same color.\n\n2. Build an adjacency matrix:\nThe build_adjacency_matrix function constructs an adjacency matrix of size R x R, where R is the number of valid rows. The matrix represents the relationship between valid rows, with adj_matrix[i][j] being True if row i and row j can be adjacent in the grid, following the problem\'s constraints. The matrix will be used in the dynamic programming step to find the number of ways to paint the grid.\n\n3. Dynamic programming:\nThe dynamic programming approach is used to count the number of ways to paint the grid. We initialize a list dp of size R, where each entry represents the number of ways to paint the grid with that specific row at the last row of the grid. Initially, we set all entries in dp to 1, as there is only one way to paint the grid with a single row.\n\nThen, we iterate through the columns (n - 1 times) and update the dp list using the adjacency matrix information. For each valid row i, we check all valid adjacent rows j (using the adjacency matrix) and update the dp list accordingly. We keep track of the number of ways to paint the grid by taking the sum of the dp list at the end and applying the modulo operation.\n\n# Complexity\n- Time complexity:\nTime Complexity:\nThe time complexity of this solution is O(R^2 * n), where R is the number of valid rows. The majority of the time is spent on the dynamic programming step, where we iterate n - 1 times, and for each iteration, we perform R * R operations when updating the dp list.\n\n- Space complexity:\nThe space complexity of the solution is O(R^2 + R * n), which is mainly due to the adjacency matrix and the dynamic programming list dp. We also store a list of valid rows, but the overall space complexity is determined by the adjacency matrix and the dp list.\n\n# More \n\nMore of my LeetCode solutions at https://github.com/aurimas13/Solutions-To-Problems.\n\n# Code\n```\nclass Solution:\n    def colorTheGrid(self, m: int, n: int) -> int:\n        MOD = 1000000007\n        valid_rows = self.generate_valid_rows(m)\n        adj_matrix = self.build_adjacency_matrix(valid_rows)\n        dp = [1] * len(valid_rows)\n        \n        for _ in range(n - 1):\n            dp_next = [0] * len(valid_rows)\n            for i, row in enumerate(valid_rows):\n                for j, adj_row in enumerate(valid_rows):\n                    if adj_matrix[i][j]:\n                        dp_next[j] += dp[i]\n                        dp_next[j] %= MOD\n            dp = dp_next\n        \n        return sum(dp) % MOD\n\n    def generate_valid_rows(self, m: int) -> list:\n        valid_rows = []\n        for i in range(3 ** m):\n            row = []\n            for j in range(m):\n                row.append((i // (3 ** j)) % 3)\n            if all(row[k] != row[k + 1] for k in range(m - 1)):\n                valid_rows.append(tuple(row))\n        return valid_rows\n\n    def build_adjacency_matrix(self, valid_rows: list) -> list:\n        adj_matrix = [[False] * len(valid_rows) for _ in range(len(valid_rows))]\n        for i, row1 in enumerate(valid_rows):\n            for j, row2 in enumerate(valid_rows):\n                if all(row1[k] != row2[k] for k in range(len(row1))):\n                    adj_matrix[i][j] = True\n        return adj_matrix\n\n```
['Dynamic Programming', 'Graph', 'Python3']