<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) {
        //similar to rotten oranges
        int n=mat.size();
        int m=mat[0].size();
        int curr=0;
        vector<vector<int>>ans(n,vector<int>(m,INT_MAX));
        vector<vector<int>>visited(n,vector<int>(m,0));
        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});
                    ans[i][j]=0;
                }
            }
        }
        int dis=0;
        while(!q.empty()){
            dis++;
            int sz=q.size();
            for(int i=0;i<sz;i++){
                auto temp=q.front();
                q.pop();
                int x=temp.first,y=temp.second;
                if(visited[x][y]==1){
                    continue;
                    //visiting at a later time step than before does not help
                }
                visited[x][y]=1;
                vector<int>dir={1,0,-1,0,1};
                for(int i=0;i<4;i++){
                    int xx=x+dir[i],yy=y+dir[i+1];
                    if(xx<0 || yy<0 || xx>=n || yy>=m)
                        continue;
                    if( visited[xx][yy])
                        continue;
                    if(mat[xx][yy]==1){
                        ans[xx][yy]=min(dis,ans[xx][yy]);
                        q.push({xx,yy});
                    }
                }
            }
        }
        return ans;
    }
};
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