<filename>773 Sliding Puzzle.py
#!/usr/bin/python3
"""
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.

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].
"""
from typing import List
from collections import defaultdict
from copy import deepcopy
import heapq


final_pos = {
    1: (0, 0),
    2: (0, 1),
    3: (0, 2),
    4: (1, 0),
    5: (1, 1),
    0: (1, 2),
}

dirs = (
    (0, -1),
    (0, 1),
    (-1, 0),
    (1, 0),
)


class Solution:
    def slidingPuzzle(self, board: List[List[int]]) -> int:
        """
        BFS + visited

        => A*
        priority = current_dist + heuristic_dist

        Chain the matrix into 1d array. N = R * C
        Complexity O(N * N!)
        There are O(N!) possible board states. O(N) is the time to scan the board for the operations in the loop.
        """
        visited = defaultdict(bool)
        m, n = len(board), len(board[0])
        q = [(self.heuristic_dist(board) + 0, 0, board)]
        target = [
            [1, 2, 3],
            [4, 5, 0],
        ]
        while q:
            heu, cur_dist, board = heapq.heappop(q)
            visited[self.ser(board)] = True
            if board == target:
                return cur_dist

            cur_dist += 1
            i, j = self.zero_pos(board)
            for di, dj in dirs:
                I = i + di
                J = j + dj
                if 0 <= I < m and 0 <= J < n:
                    B = deepcopy(board)   # need a copy in the queue
                    B[I][J], B[i][j] = B[i][j], B[I][J]
                    if not visited[self.ser(B)]:
                        heapq.heappush(q, (self.heuristic_dist(B) + cur_dist, cur_dist, B))
        return -1

    def zero_pos(self, board):
        for i, row in enumerate(board):
            for j, v in enumerate(row):
                if v == 0:
                    return i, j
        raise

    def heuristic_dist(self, board):
        """
        manhattan distance
        """
        ret = 0
        for i, row in enumerate(board):
            for j, v in enumerate(row):
                if v != 0:
                    I, J = final_pos[v]
                    ret += abs(i - I) + abs(j - J)
        return ret

    def ser(self, board):
        return tuple(
            tuple(row)
            for row in board
        )


if __name__ == "__main__":
    assert Solution().slidingPuzzle([[1,2,3],[4,0,5]]) == 1
    assert Solution().slidingPuzzle([[1,2,3],[5,4,0]]) == -1

<filename>773 Sliding Puzzle.py
#!/usr/bin/python3
"""
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.

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].
"""
from typing import List
from collections import defaultdict
from copy import deepcopy
import heapq


final_pos = {
    1: (0, 0),
    2: (0, 1),
    3: (0, 2),
    4: (1, 0),
    5: (1, 1),
    0: (1, 2),
}

dirs = (
    (0, -1),
    (0, 1),
    (-1, 0),
    (1, 0),
)


class Solution:
    def slidingPuzzle(self, board: List[List[int]]) -> int:
        """
        BFS + visited

        => A*
        priority = current_dist + heuristic_dist

        Chain the matrix into 1d array. N = R * C
        Complexity O(N * N!)
        There are O(N!) possible board states. O(N) is the time to scan the board for the operations in the loop.
        """
        visited = defaultdict(bool)
        m, n = len(board), len(board[0])
        q = [(self.heuristic_dist(board) + 0, 0, board)]
        target = [
            [1, 2, 3],
            [4, 5, 0],
        ]
        while q:
            heu, cur_dist, board = heapq.heappop(q)
            visited[self.ser(board)] = True
            if board == target:
                return cur_dist

            cur_dist += 1
            i, j = self.zero_pos(board)
            for di, dj in dirs:
                I = i + di
                J = j + dj
                if 0 <= I < m and 0 <= J < n:
                    B = deepcopy(board)   # need a copy in the queue
                    B[I][J], B[i][j] = B[i][j], B[I][J]
                    if not visited[self.ser(B)]:
                        heapq.heappush(q, (self.heuristic_dist(B) + cur_dist, cur_dist, B))
        return -1

    def zero_pos(self, board):
        for i, row in enumerate(board):
            for j, v in enumerate(row):
                if v == 0:
                    return i, j
        raise

    def heuristic_dist(self, board):
        """
        manhattan distance
        """
        ret = 0
        for i, row in enumerate(board):
            for j, v in enumerate(row):
                if v != 0:
                    I, J = final_pos[v]
                    ret += abs(i - I) + abs(j - J)
        return ret

    def ser(self, board):
        return tuple(
            tuple(row)
            for row in board
        )


if __name__ == "__main__":
    assert Solution().slidingPuzzle([[1,2,3],[4,0,5]]) == 1
    assert Solution().slidingPuzzle([[1,2,3],[5,4,0]]) == -1

#!/usr/bin/python3 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.  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]. BFS + visited          => A*         priority = current_dist + heuristic_dist          Chain the matrix into 1d array. N = R * C         Complexity O(N * N!)         There are O(N!) possible board states. O(N) is the time to scan the board for the operations in the loop. # need a copy in the queue manhattan distance