Maximum Number of Moves to Capture All Pawns on a Chessboard | Explanation
Intuition\n\nThe problem involves a knight on a chessboard trying to capture all given pawns. The knight must move optimally to maximize the number of captures 
# Intuition\n\nThe problem involves a knight on a chessboard trying to capture all given pawns. The knight must move optimally to maximize the number of captures while avoiding redundant moves. Alice and Bob alternate turns, where Alice aims to maximize the total number of moves, and Bob aims to minimize it. To solve this problem, we need to compute the minimum number of moves required to capture all pawns, considering optimal play strategies.\n\n# Approach\n\n1. **Modeling the Problem:**\n   - **Knight\'s Moves:** A knight in chess moves in an "L" shape: two squares in one direction and then one square perpendicular, or one square in one direction and two squares perpendicular. There are up to 8 possible moves.\n\n2. **Distance Calculation:**\n   - **From Knight to Pawns:** Use Breadth-First Search (BFS) to calculate the shortest distance from the knight\'s starting position to each pawn. This will help in determining the minimum number of moves needed for each capture.\n   - **Between Pawns:** Calculate the shortest distance between each pair of pawns using BFS. This information is used to compute the optimal sequence of captures.\n\n3. **Dynamic Programming (DP):**\n   - **State Representation:** Use a bitmask to represent the set of captured pawns. The DP table `dp[k][curr]` stores the maximum number of moves starting from position `k` with the captured pawns represented by the bitmask `curr`.\n   - **Recursion:** Implement a recursive function to explore all possible sequences of captures, alternating between maximizing and minimizing the total moves.\n\n4. **Minimax Strategy:**\n   - Use a minimax approach to determine the optimal sequence of captures. Alice tries to maximize the number of moves, while Bob tries to minimize them. The DP table is updated based on these decisions.\n\n5. **BFS Implementation:**\n   - **From Knight to Pawns:** Initialize the BFS queue with the knight\'s position and calculate the minimum moves required to reach each pawn.\n   - **Between Pawns:** Use BFS for each pawn to compute distances to all other pawns.\n\n# Example Illustration\n\nConsider a chessboard with a knight at position (0, 0) and three pawns at positions (1, 2), (2, 1), and (3, 3). We want to determine the optimal number of moves to capture all pawns.\n\n1. **Distance Calculation:**\n   - **From Knight to Pawns:**\n     - Use BFS from (0, 0) to calculate the shortest path to each pawn.\n     - Example distances: Knight to (1, 2) = 1 move, Knight to (2, 1) = 2 moves, Knight to (3, 3) = 3 moves.\n\n   - **Between Pawns:**\n     - Compute the shortest distance between each pair of pawns using BFS.\n     - Example distances: Pawn (1, 2) to (2, 1) = 1 move, Pawn (2, 1) to (3, 3) = 2 moves.\n\n2. **Dynamic Programming (DP) Setup:**\n   - Create a DP table to store results for each state.\n   - **Bitmask Representation:** Use bitmasks to represent which pawns have been captured.\n\n3. **Minimax Calculation:**\n   - Recursively calculate the maximum and minimum number of moves using the DP table.\n   - **Example:** If it\'s Alice\'s turn, she will choose the capture that maximizes the total moves. If it\'s Bob\'s turn, he will choose the capture that minimizes the total moves.\n\n# Complexity\n\n- **Time Complexity:** $$O(2^n \\cdot n)$$\n  - The time complexity arises from the DP table, which has \\(2^n\\) possible states (where \\(n\\) is the number of pawns). For each state, distances between pawns need to be evaluated, resulting in \\(O(n^2)\\) operations.\n\n- **Space Complexity:** $$O(2^n \\cdot n)$$\n  - The space complexity is determined by the DP table, which stores results for \\(2^n\\) states, with each state involving storage for \\(n\\) positions.\n\n# Code\n\n```java\nimport java.util.*;\n\nclass Solution {\n\n    public int maxMoves(int kx, int ky, int[][] positions) {\n        HashMap<List<Integer>, Integer> map = new HashMap<>();\n        for (int i = 0; i < positions.length; i++) {\n            map.put(List.of(positions[i][0], positions[i][1]), i);\n        }\n        int[][] d = new int[positions.length + 1][positions.length + 1];\n        maxMoves(List.of(kx, ky), map, d[positions.length]);\n        for (int i = 0; i < positions.length; i++) {\n            maxMoves(List.of(positions[i][0], positions[i][1]), map, d[i]);\n        }\n        return maxMoves(positions.length, 0, d, true, new int[positions.length + 1][1 << positions.length]);\n    }\n\n    private void maxMoves(List<Integer> list, HashMap<List<Integer>, Integer> map, int[] d) {\n        ArrayDeque<List<Integer>> deque = new ArrayDeque<>(List.of(list));\n        HashSet<List<Integer>> set = new HashSet<>();\n        for (int i = 0, c = d.length; c > 1; i++) {\n            for (int j = deque.size(); j > 0; j--) {\n                List<Integer> poll = deque.poll();\n                if (!set.contains(poll)) {\n                    set.add(poll);\n                    if (map.containsKey(poll)) {\n                        d[map.get(poll)] = i;\n                        c--;\n                    }\n                    if (poll.get(0) >= 0 && poll.get(0) < 50 && poll.get(1) >= 0 && poll.get(1) < 50) {\n                        for (int[] k : new int[][] { { -2, -1 }, { -2, 1 }, { -1, -2 }, { -1, 2 }, { 1, -2 }, { 1, 2 }, { 2, -1 }, { 2, 1 } }) {\n                            deque.offer(List.of(poll.get(0) + k[0], poll.get(1) + k[1]));\n                        }\n                    }\n                }\n            }\n        }\n    }\n\n    private int maxMoves(int k, int curr, int[][] d, boolean flag, int[][] dp) {\n        if (curr + 1 == 1 << d.length - 1) {\n            return 0;\n        } else if (dp[k][curr] == 0) {\n            dp[k][curr] = flag ? 0 : Integer.MAX_VALUE;\n            for (int i = 0; i < d.length - 1; i++) {\n                if ((1 << i & curr) == 0) {\n                    int next = maxMoves(i, 1 << i | curr, d, !flag, dp) + d[k][i];\n                    dp[k][curr] = flag ? Math.max(dp[k][curr], next) : Math.min(dp[k][curr], next);\n                }\n            }\n        }\n        return dp[k][curr];\n    }\n}\n
['Array', 'Hash Table', 'Dynamic Programming', 'Graph', 'Queue', 'Java']