Trying to understand these solution help me !!
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nimport java.util.*;\n\n// The Solution class for computing the maximum number of moves\npublic class Solution {\n    // Array to store shortest distances between pairs of positions\n    private int[][] moves = new int[17][17];\n    // Array to store dynamic programming results\n    private int[][] dp = new int[17][32767];\n    // Knight\'s movement offsets\n    private int[] knightOffsets = {1, 2, 1, -2, -1, 2, -1, -2, 1};\n\n    // Method to perform BFS and calculate shortest path distances\n    private void computeShortestPaths(int positionIndex, int[][] positions) {\n        // Initialize a distance grid with all distances set to 0\n        int[][] distanceGrid = new int[50][50];\n        int startX = positions[positionIndex][0];\n        int startY = positions[positionIndex][1];\n\n        // Queue for BFS, starting from the initial position\n        Queue<int[]> queue = new LinkedList<>();\n        queue.add(new int[] {startX, startY});\n        distanceGrid[startX][startY] = 1; // Mark the starting position as visited\n\n        while (!queue.isEmpty()) {\n            Queue<int[]> nextQueue = new LinkedList<>();\n            while (!queue.isEmpty()) {\n                int[] current = queue.poll();\n                int x = current[0];\n                int y = current[1];\n\n                // Explore all possible knight moves\n                for (int d = 0; d < 8; d++) {\n                    int newX = x + knightOffsets[d];\n                    int newY = y + knightOffsets[d + 1];\n                    \n                    // Check if the new position is within bounds and not the starting position\n                    if (newX >= 0 && newY >= 0 && newX < 50 && newY < 50 && !(newX == startX && newY == startY) && distanceGrid[newX][newY] == 0) {\n                        distanceGrid[newX][newY] = distanceGrid[x][y] + 1;\n                        nextQueue.add(new int[] {newX, newY});\n                    }\n                }\n            }\n            queue = nextQueue;\n        }\n\n        // Update the moves matrix with shortest path distances between the current position and others\n        for (int j = positionIndex + 1; j < positions.length; ++j) {\n            int targetX = positions[j][0];\n            int targetY = positions[j][1];\n            moves[positionIndex][j] = moves[j][positionIndex] = distanceGrid[targetX][targetY] - 1;\n        }\n    }\n\n    // Recursive DFS with memoization to find the optimal path score\n    private int dfs(int currentPosition, int visitedMask, int turnCount, int totalPositions) {\n        if (turnCount == totalPositions) return 0; // All positions visited\n\n        if (dp[currentPosition][visitedMask] == 0) {\n            dp[currentPosition][visitedMask] = (turnCount % 2 == 0) ? 0 : Integer.MAX_VALUE;\n\n            // Try moving to all unvisited positions\n            for (int nextPosition = 0; nextPosition < totalPositions; ++nextPosition) {\n                if ((visitedMask & (1 << nextPosition)) == 0) { // Check if nextPosition is not visited\n                    int resultFromNext = moves[currentPosition][nextPosition] + dfs(nextPosition, visitedMask | (1 << nextPosition), turnCount + 1, totalPositions);\n                    if (turnCount % 2 == 0) {\n                        dp[currentPosition][visitedMask] = Math.max(dp[currentPosition][visitedMask], resultFromNext);\n                    } else {\n                        dp[currentPosition][visitedMask] = Math.min(dp[currentPosition][visitedMask], resultFromNext);\n                    }\n                }\n            }\n        }\n\n        return dp[currentPosition][visitedMask];\n    }\n\n    // Main method to initialize data and compute the result\n    public int maxMoves(int kx, int ky, int[][] positions) {\n        int n = positions.length;\n        // Add the starting position to the list of positions\n        int[][] allPositions = new int[n + 1][2];\n        System.arraycopy(positions, 0, allPositions, 0, n);\n        allPositions[n] = new int[] {kx, ky};\n\n        // Compute shortest paths between all pairs of positions\n        for (int i = 0; i < allPositions.length; ++i) {\n            computeShortestPaths(i, allPositions);\n        }\n\n        // Compute the maximum number of moves starting from the last position with no positions visited\n        return dfs(allPositions.length - 1, 0, 0, allPositions.length - 1);\n    }\n}\n\n```
['Java']