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773. Sliding Puzzle

Description

On an 2 x 3 board, there are five tiles labeled from 1 to 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 the puzzle board 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]]

 

Constraints:

  • board.length == 2
  • board[i].length == 3
  • 0 <= board[i][j] <= 5
  • Each value board[i][j] is unique.

Solutions

Solution 1

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class Solution:
    def slidingPuzzle(self, board: List[List[int]]) -> int:
        t = [None] * 6

        def gets():
            for i in range(2):
                for j in range(3):
                    t[i * 3 + j] = str(board[i][j])
            return ''.join(t)

        def setb(s):
            for i in range(2):
                for j in range(3):
                    board[i][j] = int(s[i * 3 + j])

        def f():
            res = []
            i, j = next((i, j) for i in range(2) for j in range(3) if board[i][j] == 0)
            for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
                x, y = i + a, j + b
                if 0 <= x < 2 and 0 <= y < 3:
                    board[i][j], board[x][y] = board[x][y], board[i][j]
                    res.append(gets())
                    board[i][j], board[x][y] = board[x][y], board[i][j]
            return res

        start = gets()
        end = "123450"
        if start == end:
            return 0
        vis = {start}
        q = deque([(start)])
        ans = 0
        while q:
            ans += 1
            for _ in range(len(q)):
                x = q.popleft()
                setb(x)
                for y in f():
                    if y == end:
                        return ans
                    if y not in vis:
                        vis.add(y)
                        q.append(y)
        return -1
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class Solution {
    private String[] t = new String[6];
    private int[][] board;

    public int slidingPuzzle(int[][] board) {
        this.board = board;
        String start = gets();
        String end = "123450";
        if (end.equals(start)) {
            return 0;
        }
        Set<String> vis = new HashSet<>();
        Deque<String> q = new ArrayDeque<>();
        q.offer(start);
        vis.add(start);
        int ans = 0;
        while (!q.isEmpty()) {
            ++ans;
            for (int n = q.size(); n > 0; --n) {
                String x = q.poll();
                setb(x);
                for (String y : next()) {
                    if (y.equals(end)) {
                        return ans;
                    }
                    if (!vis.contains(y)) {
                        vis.add(y);
                        q.offer(y);
                    }
                }
            }
        }
        return -1;
    }

    private String gets() {
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 3; ++j