Leetcode: Sliding Puzzle

Sliding Puzzle



Similar Problems:


On a 2×3 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].

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/sliding-puzzle
## Basic Ideas: BFS: find the mininum path from point1 to point2
##
## Complexity:
class Solution(object):
    def slidingPuzzle(self, board):
        """
        :type board: List[List[int]]
        :rtype: int
        """
        target = '123450'
        queue, visited = [], set()
        queue.append(self.toString(board))

        level = 0
        while len(queue) != 0:
            for i in range(len(queue)):
                state = queue[0]
                if state == target: return level
                del queue[0]
                visited.add(state)
                # move to next
                self.addNeighbors(state, visited, queue)
                # print queue
            level += 1
        return -1

    def toString(self, board):
        res = ''
        for i in range(2):
            for j in range(3):
                res = res + str(board[i][j])
        return res

    def addNeighbors(self, state, visited, queue):
        # change back state to board
        matrix = [[None for j in range(3)] for i in range(2)]
        index = 0
        i0, j0 = None, None
        for i in range(2):
            for j in range(3):
                matrix[i][j] = state[index]
                index += 1
                if matrix[i][j] == '0':
                    i0, j0 = i, j
        # get next: 
        for (ik, jk) in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
            i2, j2 = i0+ik, j0+jk
            if i2<0 or i2 >= 2 or j2<0 or j2>=3: continue
            matrix[i0][j0], matrix[i2][j2] = matrix[i2][j2], matrix[i0][j0]
            newState = self.toString(matrix)
            if newState not in visited:
                queue.append(newState)
            # change back
            matrix[i0][j0], matrix[i2][j2] = matrix[i2][j2], matrix[i0][j0]
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