How do you solve Minimum Moves to Reach Target with Rotations on LeetCode?

Minimum Moves to Reach Target with Rotations (LeetCode #1210) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The snake's orientation adds complexity to the problem, requiring careful management of its state.

What is the brute force solution for Minimum Moves to Reach Target with Rotations?

The brute force approach for Minimum Moves to Reach Target with Rotations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach explores all possible paths the snake can take to reach the target. It uses a recursive method to try every possible move and rotation until it finds the target or exhausts all options. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Minimum Moves to Reach Target with Rotations?

Minimum Moves to Reach Target with Rotations uses the following concepts: Array, Breadth-First Search, Matrix. The recommended patterns to study are: Breadth-First Search, State Management.

What are the interview tips for Minimum Moves to Reach Target with Rotations?

Clearly explain your thought process and how you plan to explore the grid. Consider edge cases, such as blocked starting or ending positions. Practice BFS problems to become comfortable with queue management and state tracking.

What are common mistakes when solving Minimum Moves to Reach Target with Rotations?

Failing to account for the snake's orientation when checking valid moves. Not using a set to track visited states, leading to infinite loops.

#1210

Minimum Moves to Reach Target with Rotations

Hard

ArrayBreadth-First SearchMatrixBreadth-First SearchState Management

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal solution uses Breadth-First Search (BFS) to explore the grid efficiently. It tracks the snake's position and orientation, ensuring that we only explore valid moves and rotations, minimizing the number of moves needed to reach the target.

⚙️

Algorithm

4 steps

1Step 1: Initialize a queue for BFS with the starting position and orientation of the snake.
2Step 2: Use a set to track visited states (position and orientation) to avoid cycles.
3Step 3: For each position, explore all valid moves and rotations, adding them to the queue with an incremented move count.
4Step 4: If the target position is reached, return the number of moves; if the queue is exhausted without finding the target, return -1.

solution.py34 lines

1from collections import deque
2
3def minMoves(grid):
4    n = len(grid)
5    if grid[0][0] == 1 or grid[n-1][n-1] == 1: return -1
6    queue = deque([(0, 0, 0)])  # (row, col, orientation)
7    visited = set((0, 0, 0))
8    moves = 0
9    while queue:
10        for _ in range(len(queue)):
11            r, c, o = queue.popleft()
12            if (r, c) == (n-1, n-2) and o == 0: return moves
13            # Move right
14            if c + 1 < n and grid[r][c + 1] == 0:
15                if o == 0 and (r, c + 1, 0) not in visited:
16                    visited.add((r, c + 1, 0))
17                    queue.append((r, c + 1, 0))
18            # Move down
19            if r + 1 < n and grid[r + 1][c] == 0:
20                if o == 1 and (r + 1, c, 1) not in visited:
21                    visited.add((r + 1, c, 1))
22                    queue.append((r + 1, c, 1))
23            # Rotate clockwise
24            if o == 0 and r + 1 < n and grid[r + 1][c] == 0 and grid[r + 1][c + 1] == 0:
25                if (r, c, 1) not in visited:
26                    visited.add((r, c, 1))
27                    queue.append((r, c, 1))
28            # Rotate counterclockwise
29            if o == 1 and c + 1 < n and grid[r][c + 1] == 0 and grid[r + 1][c + 1] == 0:
30                if (r, c, 0) not in visited:
31                    visited.add((r, c, 0))
32                    queue.append((r, c, 0))
33        moves += 1
34    return -1

ℹ

Complexity note: The complexity is O(n²) due to the BFS exploring the grid, but we also store visited states which can grow with the number of unique positions and orientations.

1The snake's orientation adds complexity to the problem, requiring careful management of its state.
2BFS is particularly effective for shortest path problems in unweighted grids.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.