<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 dx[4] = {0, 0, 1, -1};
        int dy[4] = {1, -1, 0, 0};

        int n = mat.size(), m = mat[0].size();

        vector<vector<int>> dis(n, vector<int>(m,INT_MAX));

        vector<vector<bool>> vis(n, vector<bool>(m, false));

        queue<array<int, 3>> q;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (mat[i][j] == 0) {
                    dis[i][j] = 0;
                    q.push({i, j, 0});
                }
            }
        }

        while (!q.empty()) {
            auto curr = q.front();
            q.pop();

            int row = curr[0], col = curr[1], d = curr[2];

            vis[row][col] = true;

            for (int i = 0; i < 4; i++) {
                int r = dx[i] + row;
                int c = dy[i] + col;

                if (r >= 0 && c >= 0 && r < n && c < m && !vis[r][c] &&
                    mat[r][c] == 1) {
                    dis[r][c] = min(1 + d,dis[r][c]);
                    q.push({r, c, dis[r][c]});
                }
            }
        }

        return dis;
    }
};
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