<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>

#include <vector>
#include <queue>
#include <set>

using namespace std;

class Solution {
public:
    // Directions for moving: up, right, down, left
    int dx[4] = {-1, 0, 1, 0};
    int dy[4] = {0, 1, 0, -1};
    vector<vector<int>> target = {{1, 2, 3}, {4, 5, 0}}; // Target configuration

    // Function to check if a position is within the bounds of the 2x3 grid
    bool inBounds(int x, int y) {
        return x >= 0 && x < 2 && y >= 0 && y < 3;
    }

    int slidingPuzzle(vector<vector<int>>& board) {
        queue<pair<vector<vector<int>>, pair<int, pair<int, int>>>> q; // BFS queue storing board state and position of 0
        set<vector<vector<int>>> visited; // Set to track visited board states
        
        // Find the initial position of the empty space (0)
        int startRow = -1, startCol = -1;
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 3; j++) {
                if (board[i][j] == 0) {
                    startRow = i;
                    startCol = j;
                }
            }
        }
        
        // Push the initial state into the queue
        q.push({board, {0, {startRow, startCol}}});
        visited.insert(board);

        // BFS Loop
        while (!q.empty()) {
            auto [currBoard, pos] = q.front();
            int steps = pos.first;
            auto zeroPos = pos.second;
            q.pop();

            // Check if the current state matches the target
            if (currBoard == target) {
                return steps;
            }

            // Explore all four possible moves for the empty space (0)
            for (int k = 0; k < 4; k++) {
                int newRow = zeroPos.first + dx[k];
                int newCol = zeroPos.second + dy[k];

                // Check if the new position is within bounds
                if (inBounds(newRow, newCol)) {
                    // Create a new board state by swapping the 0 with the adjacent position
                    vector<vector<int>> newBoard = currBoard;
                    swap(newBoard[zeroPos.first][zeroPos.second], newBoard[newRow][newCol]);

                    // If this state has not been visited, add it to the queue and mark it as visited
                    if (!visited.count(newBoard)) {
                        visited.insert(newBoard);
                        q.push({newBoard, {steps + 1, {newRow, newCol}}});
                    }
                }
            }
        }

        // If we exhaust the queue without finding the target state, return -1
        return -1;
    }
};

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