<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:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int dirs[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
        int m = mat.size();
        int n = mat[0].size();
        vector<vector<int>> matrix(m, vector<int>(n, 0));
        vector<vector<bool>> seen(m, vector<bool>(n, false));
        queue<vector<int>> queue;

        for (int row = 0; row < m; row++) {
            for (int col = 0; col < n; col++) {
                matrix[row][col] = mat[row][col];
                if (matrix[row][col] == 0) {
                    queue.push({row, col, 0});
                    seen[row][col] = true;
                }
            }
        }

        while (!queue.empty()) {
            vector<int> curr = queue.front();
            queue.pop();

            for (auto& dir : dirs) {
                int nextRow = curr[0] + dir[0], nextCol = curr[1] + dir[1];
                if (0 <= nextRow && nextRow < m && 0 <= nextCol && nextCol < n && !seen[nextRow][nextCol]) {
                    seen[nextRow][nextCol] = true;
                    queue.push({nextRow, nextCol, curr[2] + 1});
                    matrix[nextRow][nextCol] = curr[2] + 1;
                }
            }
        }
        return matrix;
    }
};
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