Leetcode #1210: Minimum Moves to Reach Target with Rotations
In this guide, we solve Leetcode #1210 Minimum Moves to Reach Target with Rotations in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
In an n*n grid, there is a snake that spans 2 cells and starts moving from the top left corner at (0, 0) and (0, 1). The grid has empty cells represented by zeros and blocked cells represented by ones.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Breadth-First Search, Array, Matrix
Intuition
We need level-by-level exploration or shortest steps, which is ideal for BFS.
A queue naturally models the frontier of the search.
Approach
Push initial nodes into a queue and expand in layers.
Track visited nodes to prevent cycles.
Steps:
- Initialize queue with start nodes.
- Process level by level.
- Track visited nodes.
Example
Input: grid = [[0,0,0,0,0,1],
[1,1,0,0,1,0],
[0,0,0,0,1,1],
[0,0,1,0,1,0],
[0,1,1,0,0,0],
[0,1,1,0,0,0]]
Output: 11
Explanation:
One possible solution is [right, right, rotate clockwise, right, down, down, down, down, rotate counterclockwise, right, down].
Python Solution
class Solution:
def minimumMoves(self, grid: List[List[int]]) -> int:
def move(i1, j1, i2, j2):
if 0 <= i1 < n and 0 <= j1 < n and 0 <= i2 < n and 0 <= j2 < n:
a, b = i1 * n + j1, i2 * n + j2
status = 0 if i1 == i2 else 1
if (a, status) not in vis and grid[i1][j1] == 0 and grid[i2][j2] == 0:
q.append((a, b))
vis.add((a, status))
n = len(grid)
target = (n * n - 2, n * n - 1)
q = deque([(0, 1)])
vis = {(0, 0)}
ans = 0
while q:
for _ in range(len(q)):
a, b = q.popleft()
if (a, b) == target:
return ans
i1, j1 = a // n, a % n
i2, j2 = b // n, b % n
# 尝试向右平移（保持身体水平/垂直状态）
move(i1, j1 + 1, i2, j2 + 1)
# 尝试向下平移（保持身体水平/垂直状态）
move(i1 + 1, j1, i2 + 1, j2)
# 当前处于水平状态，且 grid[i1 + 1][j2] 无障碍，尝试顺时针旋转90°
if i1 == i2 and i1 + 1 < n and grid[i1 + 1][j2] == 0:
move(i1, j1, i1 + 1, j1)
# 当前处于垂直状态，且 grid[i2][j1 + 1] 无障碍，尝试逆时针旋转90°
if j1 == j2 and j1 + 1 < n and grid[i2][j1 + 1] == 0:
move(i1, j1, i1, j1 + 1)
ans += 1
return -1
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.