<p>Given an <code>m x n</code> binary matrix <code>mat</code>, return <em>the distance of the nearest </em><code>0</code><em> for each cell</em>.</p>

<p>The distance between two adjacent cells is <code>1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-1-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[0,0,0]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[0,0,0]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-2-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[1,1,1]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[1,2,1]]
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one <code>0</code> in <code>mat</code>.</li>
</ul>

#define FOR(i, start, end) for (int i = start; i < end; ++i)

class Cell {
public:
    int row, col;
    Cell() {}
    Cell(int r, int c) : row(r), col(c) {}
};
class Solution {
    vector<vector<int>> directions = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        queue<Cell*> Q;
        int n = mat.size(), m = mat[0].size();
        auto inRange = [&](int r, int c) {
            return (r >= 0 && r < n && c >= 0 && c < m);
        };
        FOR(i, 0, n) {
            FOR(j, 0, m) {
                if (mat[i][j] == 0) {
                    Q.push(new Cell(i, j));
                } else {
                    mat[i][j] = -1;
                }
            }
        }

        int ii, jj, current_level_size;
        Cell* top;

        while (!Q.empty()) {
            current_level_size = Q.size();
            FOR(i, 0, current_level_size) {
                top = Q.front();
                for (auto& direction : directions) {
                    ii = top->row + direction[0], jj = top->col + direction[1];
                    if (inRange(ii, jj) && mat[ii][jj] < 0) {
                        Q.push(new Cell(ii, jj));
                        mat[ii][jj] = mat[top->row][top->col] + 1;
                    }
                }
                Q.pop();
            }
        }

        return mat;
    }
};
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