<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>

// https://leetcode.com/problems/shortest-path-in-binary-matrix/

static const auto s_initialize = [] { std::ios::sync_with_stdio(false); std::cin.tie(nullptr); std::cout.tie(nullptr); return nullptr; }();

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        return shortestPathBinaryMatrix_v1(grid); // [26, 34] ms, 21.06 MB.
    }

    // OtherMethods:
    //
    //    Editorial, Approach 3: A* (Advanced).
    //    O( n*log(n) ); O( n ).
    //       https://en.wikipedia.org/wiki/A*_search_algorithm
    //       https://stackoverflow.com/questions/16246026/is-a-star-guaranteed-to-give-the-shortest-path-in-a-2d-grid
    //    So long as each node's H value is less than or equal to its actual distance to End, A-Star will always find the shortest path.
    //


private:
    // method 1. BFS.
    // O( n^2 ).
    int shortestPathBinaryMatrix_v1(vector<vector<int>>& grid) {
        const int n = static_cast<int>(grid.size());
        if (0 != grid[0][0] || 0 != grid[n - 1][n - 1])
            return -1;
        if (n <= 2)
            return n;
        std::vector< std::pair<int, int> > vNext, vTmpNext;
        vNext.reserve(n << 2);
        vTmpNext.reserve(n << 2);
        //
        int tLen = 1;
        vNext.emplace_back(0, 0);
        grid[0][0] = -1;
        while (!vNext.empty()) {
            ++tLen;
            vTmpNext.clear();
            for (const auto& tCell: vNext) {
                const int jMin = std::max(0, tCell.second - 1);
                const int jMax = std::min(tCell.second + 1, n - 1);
                const int iMin = std::max(0, tCell.first - 1);
                for (int i = std::min(tCell.first + 1, n - 1); iMin <= i; --i) {
                    for (int j = jMax; jMin <= j; --j) {
                        if (n - 1 == i && n - 1 == j)
                            return tLen;
                        if (0 == grid[i][j]) {
                            grid[i][j] = -1;
                            vTmpNext.emplace_back(i, j);
                        }
                    }
                }
            }
            vNext.swap(vTmpNext);
        }
        return -1;
    }

};


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