<p>On an <code>2 x 3</code> board, there are five tiles labeled from <code>1</code> to <code>5</code>, and an empty square represented by <code>0</code>. A <strong>move</strong> consists of choosing <code>0</code> and a 4-directionally adjacent number and swapping it.</p>

<p>The state of the board is solved if and only if the board is <code>[[1,2,3],[4,5,0]]</code>.</p>

<p>Given the puzzle board <code>board</code>, return <em>the least number of moves required so that the state of the board is solved</em>. If it is impossible for the state of the board to be solved, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/slide1-grid.jpg" style="width: 244px; height: 165px;" />
<pre>
<strong>Input:</strong> board = [[1,2,3],[4,0,5]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Swap the 0 and the 5 in one move.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/slide2-grid.jpg" style="width: 244px; height: 165px;" />
<pre>
<strong>Input:</strong> board = [[1,2,3],[5,4,0]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> No number of moves will make the board solved.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/slide3-grid.jpg" style="width: 244px; height: 165px;" />
<pre>
<strong>Input:</strong> board = [[4,1,2],[5,0,3]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> 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]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>board.length == 2</code></li>
	<li><code>board[i].length == 3</code></li>
	<li><code>0 &lt;= board[i][j] &lt;= 5</code></li>
	<li>Each value <code>board[i][j]</code> is <strong>unique</strong>.</li>
</ul>

using board = std::vector<std::vector<int>>;
using state = std::string;
using node = std::tuple<int, state>;
using dist = std::unordered_map<std::string, int>;

class Solution {
private:
    const int R = 2;
    const int C = 3;
    const std::string touchstone = "123450";
    const std::vector<int> shifts = {-1, 1, 3, -3};
    bool checkBoundary(int i) { return !(i < 0 || i >= R * C); }
    bool finished = false;
    dist distance;
    std::string origin;
    std::vector<std::pair<int, std::string>> generateStates(int idx, state s) 
    {
        std::vector<std::pair<int, std::string>> acc;
        state tmp;
        for(auto shift : shifts)
          if(checkBoundary(idx + shift))
          {
              if(idx == 2 && idx + shift == 3 || 
                 idx == 3 && idx + shift == 2)
                continue;
              tmp = s;
              std::swap(tmp[idx], tmp[idx + shift]);
              acc.push_back({idx + shift, tmp});
          }
        return acc;
    }
    void bfs(std::queue<node>& q) 
    {
        while(!q.empty() && !finished)
        {
            int idx;
            state s;
            node n = q.front();
            q.pop();
            std::tie(idx, s) = n;
            if(s == origin)
            {
                finished = true;
                continue;
            }
            for(auto x : generateStates(idx, s))
            {
                if(distance.find(x.second) != 
                   distance.end())
                  continue;
                distance[x.second] = distance[s] + 1;
                q.push({x.first, x.second});
            }
        }
    };
public:
    int slidingPuzzle(vector<vector<int>>& b) 
    {
        state s = "";
        for(int r = 0; r < R; r++)
          for(int c = 0; c < C; c++)
            s += std::to_string(b[r][c])[0];
        
        origin = s;
        std::queue<node> q;
        q.push({5, touchstone});
        distance[touchstone] = 0;
        bfs(q);
        auto path = distance.find(origin); 
        return path == distance.end() ? -1 : (*path).second;
    }
};
16053