Python eaaasy solution | explained
Code\npython3 [] \nclass Solution:\n    def maxMoves(self, grid: List[List[int]]) -> int:\n        rows, cols = len(grid), len(grid[0])\n        \n        def d
# Code\n```python3 [] \nclass Solution:\n    def maxMoves(self, grid: List[List[int]]) -> int:\n        rows, cols = len(grid), len(grid[0])\n        \n        def dfs(row, col):\n            if (row, col) in memo:\n                return memo[(row, col)]\n            moves = 0\n            for r, c in [(row - 1, col + 1), (row, col + 1), (row + 1, col + 1)]:\n                if 0 <= r < rows and 0 <= c < cols and grid[r][c] > grid[row][col]:\n                    moves = max(moves, 1 + dfs(r, c))\n            memo[(row, col)] = moves\n            return moves\n        \n        memo = {}\n        max_moves = 0\n        for row in range(rows):\n            max_moves = max(max_moves, dfs(row, 0))\n        \n        return max_moves\n\n\n```\n# Approach\n* Dimensions and DFS Definition: We store the grid\'s dimensions in rows and cols. Then, we define a recursive dfs function that takes the current row and column, calculating the maximum number of moves possible from that cell.\n\n* Moving Conditions: The function checks all three possible moves (top-right, right, and bottom-right). If the move is within bounds and the next cell\'s value is strictly greater than the current cell, it continues exploring from the new cell recursively.\n\n* Memoization: We use a dictionary memo to store already-computed moves from each cell to avoid redundant calculations.\n\n* Finding Maximum Moves: We start DFS from each cell in the first column, updating the max_moves variable with the highest move count found.\n\n* Result: The function returns the maximum number of moves among all starting cells.\n\n
['Dynamic Programming', 'Matrix', 'Python3']