Maximum Moves for Knight to Capture All Pawns easy solutions with explanations
Intuition:\nThe problem involves a knight on a chessboard that needs to capture all pawns in the minimum number of moves. The knight can move in typical "L-shap
### Intuition:\nThe problem involves a knight on a chessboard that needs to capture all pawns in the minimum number of moves. The knight can move in typical "L-shaped" moves, and the goal is to compute the maximum moves required to capture all pawns starting from the knight\'s initial position.\n\n### Approach:\n1. **BFS for Knight\'s Move Calculation:** \n   We use BFS to compute the minimum distance from the knight\'s position to each pawn. BFS ensures that the shortest path is found for each knight move.\n\n2. **Dynamic Programming with Bitmasking:** \n   To track which pawns are captured, we use bitmasking. A bitmask keeps track of the state (captured or not) of the pawns, and dynamic programming (DP) helps to store and compute the maximum and minimum moves efficiently.\n   \n   - **DP State:** \n     - `dp[mask][i]` stores the number of moves required to capture all pawns given the current mask and the last captured pawn.\n     - The mask represents which pawns are captured.\n   \n   - **Recursive Transitions:**\n     - For each possible mask, we calculate the minimum and maximum moves by transitioning between pawns using BFS-calculated distances.\n\n3. **Maximizing the Moves:**\n   - We attempt to maximize the moves by trying different orders of capturing pawns.\n   - For each pawn, we try to minimize the number of moves for subsequent pawns by exploring all permutations.\n\n4. **Optimization:**\n   - We store intermediate results in the DP array to avoid recalculating moves for already visited states.\n   - We alternate between minimizing and maximizing moves to ensure the optimal solution.\n\n### Time Complexity:\n- **BFS:** For each pawn-to-pawn move, we perform a BFS, making the complexity $O(n * m)$, where `n` is the number of pawns and `m` is the number of cells on the board.\n- **DP:** There are $2^n$ possible masks, and for each, we perform transitions in $O(n)$ time, leading to a complexity of O(n * 2^n).\n\n### Space Complexity:\n- The space complexity is $O(n * 2^n)$ due to the DP table and bitmasking.\n\n\n# Code\n```python3 []\nfrom collections import deque\nfrom typing import List\n\nclass Solution(object):\n    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:\n        """\n        :type kx: int\n        :type ky: int\n        :type positions: List[List[int]]\n        :rtype: int\n        """\n        knight_moves = [(-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1)]\n        n = len(positions)\n        \n        # BFS to find shortest distance between two points on the chessboard\n        def bfs(sx: int, sy: int, ex: int, ey: int) -> int:\n            queue = deque([(sx, sy, 0)])\n            visited = [[False] * 50 for _ in range(50)]\n            visited[sx][sy] = True\n            while queue:\n                x, y, dist = queue.popleft()\n                if x == ex and y == ey:\n                    return dist\n                for dx, dy in knight_moves:\n                    nx, ny = x + dx, y + dy\n                    if 0 <= nx < 50 and 0 <= ny < 50 and not visited[nx][ny]:\n                        visited[nx][ny] = True\n                        queue.append((nx, ny, dist + 1))\n            return float(\'inf\')\n\n        # Precompute distances from the knight to all pawns and between all pawns\n        distances = [[0] * n for _ in range(n)]\n        knight_to_pawn = []\n        for i in range(n):\n            px, py = positions[i]\n            knight_to_pawn.append(bfs(kx, ky, px, py))\n            for j in range(i + 1, n):\n                px2, py2 = positions[j]\n                dist = bfs(px, py, px2, py2)\n                distances[i][j] = distances[j][i] = dist\n\n        # DP with bitmasking: dp[mask][i] = max moves starting from pawn i with pawns represented by mask\n        dp = [[-1] * n for _ in range(1 << n)]\n        \n        # Function to calculate maximum moves with dynamic programming and bitmask\n        def dp_max(mask: int, last: int) -> int:\n            if dp[mask][last] != -1:\n                return dp[mask][last]\n            \n            total_pawns = bin(mask).count(\'1\')  # Total pawns captured so far\n            if total_pawns == n:  # All pawns captured\n                return 0\n\n            max_moves = 0\n            for i in range(n):\n                if not (mask & (1 << i)):  # If pawn i is not captured yet\n                    next_mask = mask | (1 << i)  # Capture pawn i\n                    move_cost = distances[last][i]\n                    max_moves = max(max_moves, move_cost + dp_min(next_mask, i))  # Try to maximize Alice\'s move\n            dp[mask][last] = max_moves\n            return max_moves\n\n        # Function to calculate minimum moves for Bob\n        def dp_min(mask: int, last: int) -> int:\n            if dp[mask][last] != -1:\n                return dp[mask][last]\n            \n            total_pawns = bin(mask).count(\'1\')  # Total pawns captured so far\n            if total_pawns == n:  # All pawns captured\n                return 0\n\n            min_moves = float(\'inf\')\n            for i in range(n):\n                if not (mask & (1 << i)):  # If pawn i is not captured yet\n                    next_mask = mask | (1 << i)  # Capture pawn i\n                    move_cost = distances[last][i]\n                    min_moves = min(min_moves, move_cost + dp_max(next_mask, i))  # Bob tries to minimize moves\n            dp[mask][last] = min_moves\n            return min_moves\n\n        # Start DP from the knight\'s initial position\n        max_result = 0\n        for i in range(n):\n            max_result = max(max_result, knight_to_pawn[i] + dp_min(1 << i, i))\n\n        return max_result\n\n```
['Python3']