<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(),m=mat[0].size();
        vector<vector<int>> v(n, vector<int> (m,INT_MAX));
        queue<pair<pair<int,int>,int>> q;
        
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mat[i][j]==0){
                    // v[i][j]=0;
                    q.push({{i,j},0});
                }
            }
        }
        int dx[]={0,-1,0,1};
        int dy[]={-1,0,1,0};
        while(q.size()>0){
            auto p= q.front();q.pop();
            int x=p.first.first,y=p.first.second,d=p.second;
            // cout<<x<<" "<<y<<" "<<d<<endl;
            if(v[x][y]<=d)continue;
            v[x][y]=d;
            int newx,newy;
            for(int i=0;i<4;i++){
                newx=x+dx[i],newy=y+dy[i];
                if(newx>=0 && newx<n && newy>=0 && newy<m){
                    q.push({{newx,newy},d+1});
                }
            }
        }
        return v;
    }
};
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