<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 {
    size_t rows, columns;
    vector<vector<int>> distances;

    void foreachAdj(size_t r, size_t c, auto f) {
        if (r > 0)
            f(r - 1, c);
        if (c > 0)
            f(r, c - 1);
        if (r < rows - 1)
            f(r + 1, c);
        if (c < columns - 1)
            f(r, c + 1);
    }

public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        rows = mat.size();
        columns = mat[0].size();
        distances.resize(rows);
        for (auto& r : distances) {
            r.resize(columns);
            fill(begin(r), end(r), -1);
        }

        queue<pair<size_t,size_t>> q;
        for (int r = 0; r < rows; ++r) {
            for (int c = 0; c < columns; ++c) {
                if (mat[r][c] == 0) {
                    distances[r][c] = 0;
                    q.push({r, c});
                }
            }
        }

        while (!q.empty()) {
            auto [r, c] = q.front();
            q.pop();
            foreachAdj(r, c, [&] (size_t rc, size_t cc) {
                if (distances[rc][cc] >= 0)
                    return;
                distances[rc][cc] = distances[r][c] + 1;
                q.push({rc, cc});
            });
        }

        return distances;
    }
};
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