<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>

#define code ios_base::sync_with_stdio(false);
#define by cin.tie(NULL);
#define bitandas cout.tie(NULL);

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        code by bitandas;
        int m = mat.size(), n = mat[0].size();;
        vector<vector<int>> ans(m , vector<int> (n, 0));
        vector<vector<bool>> vis(m , vector<bool> (n, 0));
        queue<vector<int>> q;
        int diff[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};
        for (int i = 0; i < m; i++)  {
            for (int j = 0; j < n; j++) {
                if (!mat[i][j]) {
                    vis[i][j] = 1;
                    continue;
                }
                for (auto k : diff) {
                    int x = i + k[0], y = j + k[1];
                    if (x < 0 || x >= m || y < 0 || y >= n) continue;
                    if (!mat[x][y]) {
                        vis[i][j] = 1;
                        q.push({i, j, 1});
                        break;
                    }
                }
            }
        }

        while (q.size()) {
                auto j = q.front(); q.pop();
                ans[j[0]][j[1]] = j[2];
                for (auto k : diff) {
                    int x = j[0] + k[0], y = j[1] + k[1];
                    if (x < 0 || x >= m || y < 0 || y >= n) continue;
                    if (mat[x][y] && !vis[x][y]) {
                        vis[x][y] = 1;
                        q.push({x, y, j[2] + 1});
                    }
                }
        }
        return ans;
    }
};
48760