<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>

class Solution {
public:
    int dp[44791];
    int dfg(int val, int po, vector<int>& v, int tar) {
        queue<pair<int,int>> q;
        q.push({val, po});
        dp[val] = 0;
        int z;

        while (!q.empty()) {
            auto pr = q.front();
            val = pr.first;
            po = pr.second;
            q.pop();

            // Move Up (only if po >= 3)
            if (po >= 3) {
                z = (val / v[po - 3]) % 6;
                z = z * (v[po] - v[po - 3]);
                if (dp[z + val] == -1) {
                    dp[z + val] = dp[val] + 1;
                    if (z + val == tar) break;
                    q.push({z + val, po - 3});
                }
            }
            // Move Down (only if po <= 2)
            if (po <= 2) {
                z = (val / v[po + 3]) % 6;
                z = z * (v[po] - v[po + 3]);
                if (dp[z + val] == -1) {
                    dp[z + val] = dp[val] + 1;
                    if (z + val == tar) break;
                    q.push({z + val, po + 3});
                }
            }
            // Move Left (only if po % 3 != 0)
            if (po % 3 != 0) {
                z = (val / v[po - 1]) % 6;
                z = z * (v[po] - v[po - 1]);
                if (dp[z + val] == -1) {
                    dp[z + val] = dp[val] + 1;
                    if (z + val == tar) break;
                    q.push({z + val, po - 1});
                }
            }
            // Move Right (only if po % 3 != 2)
            if (po % 3 != 2) {
                z = (val / v[po + 1]) % 6;
                z = z * (v[po] - v[po + 1]);
                if (dp[z + val] == -1) {
                    dp[z + val] = dp[val] + 1;
                    if (z + val == tar) break;
                    q.push({z + val, po + 1});
                }
            }
        }
        return dp[tar];
    }

    int slidingPuzzle(vector<vector<int>>& b) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        memset(dp, -1, sizeof(dp));
        int val = 0, mu = 1, x = 0, po;
        vector<int> v;
        for (int i = 1; i >= 0; i--) {
            for (int j = 2; j >= 0; j--) {
                val += mu * b[i][j];
                if (b[i][j] == 0) po = x;
                v.push_back(mu);
                mu *= 6;
                x++;
            }
        }
        if (val == 11190) return 0;
        return dfg(val, po, v, 11190);
    }
};
9853