Sliding Puzzle
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0773.Sliding%20Puzzle/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0773.Sliding%20Puzzle/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0773.Sliding%20Puzzle/images/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>
Breadth-First Search; Memoization; Array; Dynamic Programming; Backtracking; Matrix
class Solution {
public:
    int slidingPuzzle(vector<vector<int>>& board) {
        string start = gets(board);
        string end = "123450";
        if (start == end) return 0;
        unordered_set<string> vis;
        vis.insert(start);
        queue<string> q{{start}};
        int ans = 0;
        while (!q.empty()) {
            ++ans;
            for (int n = q.size(); n > 0; --n) {
                string x = q.front();
                q.pop();
                setb(x, board);
                for (string y : next(board)) {
                    if (y == end) return ans;
                    if (!vis.count(y)) {
                        vis.insert(y);
                        q.push(y);
                    }
                }
            }
        }
        return -1;
    }

    string gets(vector<vector<int>>& board) {
        string s = "";
        for (int i = 0; i < 2; ++i)
            for (int j = 0; j < 3; ++j)
                s.push_back('0' + board[i][j]);
        return s;
    }

    void setb(string s, vector<vector<int>>& board) {
        for (int i = 0; i < 2; ++i)
            for (int j = 0; j < 3; ++j)
                board[i][j] = s[i * 3 + j] - '0';
    }

    vector<string> next(vector<vector<int>>& board) {
        vector<string> res;
        auto p = find0(board);
        int i = p.first, j = p.second;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < 2 && y >= 0 && y < 3) {
                swap(i, j, x, y, board);
                res.push_back(gets(board));
                swap(i, j, x, y, board);
            }
        }
        return res;
    }

    void swap(int i, int j, int x, int y, vector<vector<int>>& board) {
        int t = board[i][j];
        board[i][j] = board[x][y];
        board[x][y] = t;
    }

    pair<int, int> find0(vector<vector<int>>& board) {
        for (int i = 0; i < 2; ++i)
            for (int j = 0; j < 3; ++j)
                if (board[i][j] == 0)
                    return {i, j};
        return {0, 0};
    }
};