773. Sliding Puzzle
Category: /leetcodeOn a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square represented by 0.
A move consists of choosing 0 and a 4-directionally adjacent number and swapping it.
The state of the board is solved if and only if the board is [[1,2,3],[4,5,0]].
Given a puzzle board, return the least number of moves required so that the state of the board is solved. If it is impossible for the state of the board to be solved, return -1.
Examples:
Input: board = [[1,2,3],[4,0,5]]
Output: 1
Explanation: Swap the 0 and the 5 in one move.
Input: board = [[1,2,3],[5,4,0]]
Output: -1
Explanation: No number of moves will make the board solved.
Input: board = [[4,1,2],[5,0,3]]
Output: 5
Explanation: 5 is the smallest number of moves that solves the board.
An example path:
After move 0: [[4,1,2],[5,0,3]]
After move 1: [[4,1,2],[0,5,3]]
After move 2: [[0,1,2],[4,5,3]]
After move 3: [[1,0,2],[4,5,3]]
After move 4: [[1,2,0],[4,5,3]]
After move 5: [[1,2,3],[4,5,0]]
Input: board = [[3,2,4],[1,5,0]]
Output: 14
Note:
- board will be a 2 x 3 array as described above.
- board[i][j] will be a permutation of [0, 1, 2, 3, 4, 5].
Solution
- BFS template, need to convert 2D grid to a list to compare the target state, then have a mapping for the nerghbor nodes.
- T: O(MN(M*N)!)
- S: O(MN(M*N)!)
def slidingPuzzle(self, board):
if not board: return -1
rows, cols = len(board), len(board[0])
start = ''
# or itertools.chain(*board)
target = '123450'
# target = "".join(range(1, rows * cols) + [0])
neighbor = [{1, 3}, {0, 4, 2}, {1, 5},
{0, 4}, {3, 1, 5}, {4, 2}]
# BFS template
q = collections.deque()
visited = set()
q.append(start)
visited.add(start)
step = 0
while q:
sz = len(q)
for i in range(sz):
curr = q.popleft()
# check target state
if curr == target:
return step
# find '0' index
idx = 0
while curr[0] != '0':
idx += 1
for adj in neighbor[idx]:
# swap 0 with the next one
new_board = list(curr)
new_board[adj], new_board[idx] = new_board[idx], new_board[adj]
new_board = "".join(new_board)
if new_board not in visited:
q.append(new_board)
visited.add(new_board)
step += 1
return -1
