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

        vector<vector<int>> visited(m, vector<int> (n,0));
        vector<vector<int>> res(m, vector<int> (n,0));
        queue<pair<int,int>> q;

        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if (mat[i][j] == 0) {
                    visited[i][j] =1;
                    q.push({i,j});
                    res[i][j] = 0;
                }
            }
        }

        while (!q.empty()) {
            int x = q.front().first;
            int y = q.front().second;
            int d = res[x][y];
            vector<int> dirx = {1,0,-1,0};
            vector<int> diry = {0, 1, 0, -1};
            q.pop();

            for (int i=0; i<4; i++) {
                int xx = x+dirx[i];
                int yy = y+diry[i];
                if (xx>=0 && xx<m && yy>=0 && yy<n && visited[xx][yy] != 1) {
                    visited[xx][yy] =1;
                    res[xx][yy] = d+1;
                    q.push({xx,yy});
                }
            }
        }

        return res;
    }
};
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