<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<pair<int,int>> dir = {{1,0}, {0,1}, {-1,0}, {0,-1}};

    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int m = mat.size();
        int n = mat[0].size();
        unordered_map<int,int> visited;
        unordered_set<int> seen;
        vector<vector<int>> res;
        for (int i=0; i<m; i++) {
            res.push_back({});
            for (int j=0; j<n; j++) {
                res[i].push_back(0);
            }
        }

        queue<pair<int,int>> queue;
        int dist = 1;
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if (mat[i][j] == 0) {
                    int id = i*n+j;
                    seen.insert(id);
                    queue.push({i,j});
                }
            }
        }
        while (!queue.empty()) {
            int queue_size = queue.size();
            //cout<<"dist = "<<dist<<" size="<<queue_size<<endl;
            for (int rep=0; rep<queue_size; rep++) {
                auto [i,j] = queue.front();
                queue.pop();
                //cout<<"popping = "<<i<<" "<<j<<endl;
                for (auto [di,dj]:dir) {
                    int ni = i+di;
                    int nj = j+dj;
                    int nid = ni*n+nj;
                    if (seen.count(nid))
                        continue;
                    if (ni<0 || nj<0 || ni>=m || nj>=n)
                        continue;
                    //cout<<"pos "<<ni<<" "<<nj<<endl;
                    res[ni][nj] = dist;
                    queue.push({ni,nj});
                    seen.insert(nid);
                }
            }
            dist++;
        }
        return res;
    }
};
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