Simple Python3 solution [EXPLAINED]
Intuition\nStarting from any cell in the first column, we want to find the longest path by moving only to larger numbers in the next column. We\u2019ll track ou
# Intuition\nStarting from any cell in the first column, we want to find the longest path by moving only to larger numbers in the next column. We\u2019ll track our path length as we go, aiming to keep it as long as possible.\n\n# Approach\nUse DP with DFS to explore each cell in the first column, storing the longest path from each cell so we never repeat work.\n\n# Complexity\n- Time complexity:\nO(m * n), where we check each cell once and store results.\n\n- Space complexity:\nO(m * n), for storing path lengths for each cell in the grid.\n\n# Code\n```python3 []\nclass Solution:\n    def maxMoves(self, grid: List[List[int]]) -> int:\n        m, n = len(grid), len(grid[0])\n        dp = [[-1] * n for _ in range(m)]\n        \n        # Initialize max_moves to zero\n        max_moves = 0\n        \n        # Recursive DFS with memoization\n        def dfs(row, col):\n            # If already computed, return the stored result\n            if dp[row][col] != -1:\n                return dp[row][col]\n            \n            # Initialize the move count from this cell to 0\n            moves = 0\n            \n            # Possible moves to the next column (row-1, row, row+1)\n            for dr in [-1, 0, 1]:\n                new_row, new_col = row + dr, col + 1\n                if 0 <= new_row < m and new_col < n and grid[new_row][new_col] > grid[row][col]:\n                    moves = max(moves, 1 + dfs(new_row, new_col))\n            \n            # Memoize and return result\n            dp[row][col] = moves\n            return moves\n        \n        # Start from each cell in the first column\n        for row in range(m):\n            max_moves = max(max_moves, dfs(row, 0))\n        \n        return max_moves\n\n```
['Dynamic Programming', 'Python3']