<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 row = mat.size(), col = mat[0].size();
        queue<pair<int, int>> q;
        vector<vector<int>> vis(row, vector<int>(col, -1));
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (mat[i][j] == 0) {
                    vis[i][j] = 0;
                    q.push({i, j});
                }
            }
        }
        vector<int> drow = {-1, 0, 1, 0};
        vector<int> dcol = {0, -1, 0, 1};
        while (!q.empty()) {
            int row1 = q.front().first;
            int col1 = q.front().second;
            q.pop();
            for (int i = 0; i < 4; i++) {
                int nrow = row1 + drow[i];
                int ncol = col1 + dcol[i];
                if (nrow >= 0 && nrow < row && ncol >= 0 && ncol < col &&
                    vis[nrow][ncol] == -1) {
                    vis[nrow][ncol] = vis[row1][col1] + 1;
                    q.push({nrow, ncol});
                }
            }
        }
        return vis;
    }
};
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