LeetCode 1210: Minimum Moves to Reach Target with Rotations Solution

Master LeetCode problem 1210 (Minimum Moves to Reach Target with Rotations), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1210. Minimum Moves to Reach Target with Rotations

Hard

Array Breadth-First Search Matrix

Problem Explanation

Explanation

To solve this problem, we can use a breadth-first search (BFS) approach considering the snake's position and direction in each state. We can maintain a visited set to avoid revisiting the same state. At each step, we can move right, move down, rotate clockwise, or rotate counterclockwise if the conditions allow.

Initialize a queue with initial state (0, 0, 0, 1) representing the snake's head position, tail position, and direction.
Keep track of the minimum number of moves required to reach the target.
Perform BFS by exploring all possible moves from the current state.
Update the minimum moves when reaching the target position.
Return the minimum moves or -1 if the target is unreachable.

Time Complexity: The time complexity of the BFS approach is O(n^2) as we explore each cell once.

Space Complexity: The space complexity is O(n^2) to store the visited states.

Solution Code

class Solution {
    public int minimumMoves(int[][] grid) {
        int n = grid.length;
        Set<String> visited = new HashSet<>();
        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{0, 0, 0, 1}); // (head row, head col, tail row, tail col)

        int moves = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int[] curr = queue.poll();

                // Check if snake reaches the target
                if (curr[0] == n - 1 && curr[1] == n - 1 && curr[2] == n - 1 && curr[3] == n - 2) {
                    return moves;
                }

                String state = Arrays.toString(curr);
                if (visited.contains(state)) {
                    continue;
                }
                visited.add(state);

                // Move right
                if (curr[1] + 1 < n && grid[curr[0]][curr[1] + 1] == 0) {
                    queue.offer(new int[]{curr[0], curr[1] + 1, curr[2], curr[3] + 1});
                }

                // Move down
                if (curr[0] + 1 < n && grid[curr[0] + 1][curr[1]] == 0 && grid[curr[2] + 1][curr[3]] == 0) {
                    queue.offer(new int[]{curr[0] + 1, curr[1], curr[2] + 1, curr[3]});
                }

                // Rotate clockwise
                if (curr[0] == curr[2] && curr[1] + 1 < n && grid[curr[0] + 1][curr[1] + 1] == 0) {
                    queue.offer(new int[]{curr[0], curr[1], curr[0], curr[1] + 1});
                }

                // Rotate counterclockwise
                if (curr[1] == curr[3] && curr[0] + 1 < n && grid[curr[0] + 1][curr[1] + 1] == 0) {
                    queue.offer(new int[]{curr[0], curr[1], curr[0] + 1, curr[1]});
                }
            }
            moves++;
        }
        return -1;
    }
}

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