773. Sliding Puzzle

Problem

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.

Solution (Java)

import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;

public class Solution {
    private static class Node {
        public String board;
        public int depth;
        public int y;
        public int x;

        public Node(String board, int depth, int y, int x) {
            this.board = board;
            this.depth = depth;
            this.y = y;
            this.x = x;
        }
    }

    public int slidingPuzzle(int[][] board) {
        String targetStr = "123450";
        StringBuilder sb = new StringBuilder();
        int y = 0;
        int x = 0;
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == 0) {
                    y = i;
                    x = j;
                }
                sb.append(board[i][j]);
            }
        }
        Set<String> seen = new HashSet<>();
        Queue<Node> q = new LinkedList<>();
        q.add(new Node(sb.toString(), 0, y, x));
        int[][] dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        while (!q.isEmpty()) {
            Node next = q.poll();
            String s = next.board;
            if (!seen.contains(s)) {
                if (s.equals(targetStr)) {
                    return next.depth;
                }
                int nextDepth = next.depth + 1;
                y = next.y;
                x = next.x;
                for (int[] vector : dir) {
                    int nextY = y + vector[0];
                    int nextX = x + vector[1];
                    if (0 <= nextY
                            && nextY < board.length
                            && 0 <= nextX
                            && nextX < board[0].length) {
                        String newBoard = swap(s, y, x, nextY, nextX);
                        q.add(new Node(newBoard, nextDepth, nextY, nextX));
                    }
                }
                seen.add(s);
            }
        }
        return -1;
    }

    public String swap(String board, int y1, int x1, int y2, int x2) {
        char[] arr = board.toCharArray();
        char t = board.charAt(y1 * 3 + x1);
        arr[y1 * 3 + x1] = board.charAt(y2 * 3 + x2);
        arr[y2 * 3 + x2] = t;
        return new String(arr);
    }
}

Solution (C++)

class Solution {
public:
  int slidingPuzzle(vector<vector<int>>& board) {
    constexpr int kRows = 2;
    constexpr int kCols = 3; 
    string goal;
    string start;
    for (int i = 0; i < board.size(); ++i)
      for (int j = 0; j < board[0].size(); ++j) {
        start += (board[i][j] + '0');
        goal += (i * kCols + j + 1) % (kRows * kCols) + '0'; // 12345...0
    }

    if (start == goal) return 0;

    constexpr int dirs[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};

    set<string> visited{start};
    int steps = 0;
    queue<string> q;
    q.push(start);
    while (!q.empty()) {
      ++steps;
      int size = q.size();
      while (size-- > 0) {
        string s = q.front();
        q.pop();
        int p = s.find('0');
        int y = p / kCols;
        int x = p % kCols;        
        for (int i = 0; i < 4; ++i) {
          int tx = x + dirs[i][0];
          int ty = y + dirs[i][1];
          if (tx < 0 || ty < 0 || tx >= kCols || ty >= kRows) continue;
          int pp = ty * kCols + tx;
          string t(s);
          swap(t[p], t[pp]);          
          if (visited.count(t)) continue;            
          if (t == goal) return steps;
          visited.insert(t);
          q.push(t);
        }
      }      
    }
    return -1;
  }
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).