Sliding Puzzle
Created: February 17, 2020 by [lek-tin]
Last updated: February 17, 2020
On 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.
Example 1
Input: board = [[1,2,3],[4,0,5]]
Output: 1
Explanation: Swap the 0 and the 5 in one move.
Example 2
Input: board = [[1,2,3],[5,4,0]]
Output: -1
Explanation: No number of moves will make the board solved.
Example 3
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 3array as described above. board[i][j]will be a permutation of[0, 1, 2, 3, 4, 5].
Solution
Time: O(n*m!)
Time: O(n*m!)
from collections import deque
class Solution:
def slidingPuzzle(self, board: List[List[int]]) -> int:
n = len(board)
m = len(board[0])
start, goal = "", ""
for i in range(n):
for j in range(m):
curr = board[i][j]
if curr == 0:
start += "XX"
elif curr < 10:
start += "0" + str(curr)
else:
start += str(curr)
order = (i*m+j+1) % (n*m)
if order == 0:
goal += "XX"
elif order < 10:
goal += "0" + str(order)
else:
goal += str(order)
if start == goal:
return 0
dirs = [ [0,1], [1,0], [0,-1], [-1,0] ]
q = deque([])
visited = set()
steps = 0
q.append(start)
while q:
# increment by 1 for each layer
steps += 1
size = len(q)
# Expand layer by layer: up, down, left or right
while size > 0:
size -= 1
s = q.popleft()
p = s.index('XX')
y = (p//2) // m
x = (p//2) % m
for i in range(len(dirs)):
dx = x + dirs[i][0]
dy = y + dirs[i][1]
if dx < 0 or dy <0 or dx >= m or dy >= n:
continue
new_p = (dy*m + dx)*2
t_arr = list(s)
t_arr[p:p+2], t_arr[new_p:new_p+2] = t_arr[new_p:new_p+2], t_arr[p:p+2]
t = "".join(t_arr)
if t in visited:
continue
if t == goal:
return steps
visited.add(t)
q.append(t)
return -1
# Input: [[4,1,2],[5,0,3]]
# output: 5
# 412503
# --------
# 412530
# 402513
# 412053
# --------
# 410532
# 412503
# 420513
# 042513
# 012453
# --------
# 401532
# 423510
# 542013
# 102453
# --------
# 431502
# 041532
# 423501
# 542103
# 152403
# 120453
Improvement
We can optimize our algorithm with A*. We can use a string to represent a state and have all the trasitions pre-computed.