<p>Given an <code>n x n</code> binary matrix <code>grid</code>, return <em>the length of the shortest <strong>clear path</strong> in the matrix</em>. If there is no clear path, return <code>-1</code>.</p>

<p>A <strong>clear path</strong> in a binary matrix is a path from the <strong>top-left</strong> cell (i.e., <code>(0, 0)</code>) to the <strong>bottom-right</strong> cell (i.e., <code>(n - 1, n - 1)</code>) such that:</p>

<ul>
	<li>All the visited cells of the path are <code>0</code>.</li>
	<li>All the adjacent cells of the path are <strong>8-directionally</strong> connected (i.e., they are different and they share an edge or a corner).</li>
</ul>

<p>The <strong>length of a clear path</strong> is the number of visited cells of this path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example1_1.png" style="width: 500px; height: 234px;" />
<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example2_1.png" style="height: 216px; width: 500px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j] is 0 or 1</code></li>
</ul>

#include <iostream>
#include <vector>
#include <queue>
#include <tuple>

using namespace std;

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int n = grid.size();

        // Check if start or end is blocked
        if (grid[0][0] == 1 || grid[n-1][n-1] == 1) {
            return -1;
        }

        // 8 possible directions for movement
        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
        
        // Initialize BFS with the starting point (0,0)
        queue<tuple<int, int, int>> q;  // (row, col, distance)
        q.push({0, 0, 1});

        // Visited set to mark cells as visited
        vector<vector<bool>> visited(n, vector<bool>(n, false));
        visited[0][0] = true;

        while (!q.empty()) {
            auto [row, col, dist] = q.front();
            q.pop();

            // If we've reached the bottom-right corner
            if (row == n - 1 && col == n - 1) {
                return dist;
            }

            // Explore all 8 directions
            for (auto [dr, dc] : directions) {
                int new_row = row + dr;
                int new_col = col + dc;

                // Check if the new position is valid and not yet visited
                if (new_row >= 0 && new_row < n && new_col >= 0 && new_col < n &&
                    !visited[new_row][new_col] && grid[new_row][new_col] == 0) {

                    q.push({new_row, new_col, dist + 1});
                    visited[new_row][new_col] = true;
                }
            }
        }

        // If no path was found
        return -1;
    }
};

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