<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(const vector<vector<int>>& mat) {
        const int R = mat.size();
        const int C = mat[0].size();
        vector<vector<int>> result;
        for (int i = 0; i < R; ++i) {
            result.push_back(vector<int>(C, -1));
        }
        queue<tuple<int, int, int>> q;
        for (int r = 0; r < R; ++r) {
            for (int c = 0; c < C; ++c) {
                if (mat[r][c] == 0) {
                    q.push(make_tuple(r, c, 0));
                }
            }
        }
        while (!q.empty()) {
            auto [r, c, dist] = q.front();
            q.pop();
            if (result[r][c] == -1) {
                result[r][c] = dist;
                if (r > 0) q.push({r - 1, c, dist + 1});
                if (r + 1 < R) q.push({r + 1, c, dist + 1});
                if (c > 0) q.push({r, c - 1, dist + 1});
                if (c + 1 < C) q.push({r, c + 1, dist + 1});
            } else {
                result[r][c] = min(result[r][c], dist);
            }
        }
        return result;
    }
};
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