<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>

class Solution {
public:
    // BFS
    // starting with all the 0s
    // before enqueue, update matrix with steps
    // and mark as visited
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        queue<int> queue;
        const int rows = mat.size();
        const int cols = mat[0].size();
        vector<bool> visited(rows * cols);
        for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < cols; ++j) {
                if (mat[i][j] == 0) {
                    const int index = i * cols + j;
                    queue.push(index);
                    visited[index] = true;
                }
            }
        }

        int steps = 0;
        while (!queue.empty()) {
            for (int sz = queue.size(); sz > 0; --sz) {
                const int index = queue.front();
                queue.pop();
                const int x = index / cols;
                const int y = index % cols;
                mat[x][y] = steps;      // 0 -> 0
                constexpr int kDirs[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
                for (auto [dx, dy] : kDirs) {
                    const int nx = x + dx;
                    const int ny = y + dy;
                    if (nx < 0 || nx >= rows || ny < 0 || ny >= cols)
                        continue;
                    const int nextIndex = nx * cols + ny;
                    if (visited[nextIndex])
                        continue;
                    queue.push(nextIndex);
                    visited[nextIndex] = true;
                }
            }
            ++steps;
        }
        return mat;
    }
};
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