<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 {
    static constexpr std::pair<int,int> dirs[4] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int i, j;
        int rows, cols;
        std::queue<std::pair<int, int>> bfsQueue;

        rows = mat.size();
        cols = mat[0].size();
        for (i = 0; i < rows; i++) {
            for (j = 0; j < cols; j++) {
                if (mat[i][j] == 0) {
                    bfsQueue.emplace(i, j);
                } else {
                    mat[i][j] = -1;
                }
            }
        }

        while (!bfsQueue.empty()) {
            auto [r, c] = bfsQueue.front();
            bfsQueue.pop();

            for (const auto& [dr, dc] : dirs) {
                int newRow = r + dr;
                int newCol = c + dc;

                if ( (newRow < 0) || (newRow >= rows) || (newCol < 0) || (newCol >= cols) || (mat[newRow][newCol] != -1) ) {
                    continue;
                }

                mat[newRow][newCol] = mat[r][c] + 1;
                bfsQueue.emplace(newRow, newCol);
            }
        }

        return mat;
    }
};
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