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

        vector<vector<int>> dist (n,vector<int>(m,-1));

        queue<pair<int,int>> q;

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

        while(!q.empty()){
            auto [i,j] = q.front();
            q.pop();
            vector<pair<int,int>> dir = {{0,1},{0,-1},{1,0},{-1,0}};
            for(auto [r,c]: dir){
                int ni = i+r;
                int nj = j+c;
                if(ni<0 || nj <0 || ni >=n || nj >=m) continue;
                if(dist[ni][nj]!= -1) continue;
                dist[ni][nj]=dist[i][j]+1;
                q.push({ni,nj});
            }
        }
        return dist;
    }
};
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