<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) 
    {
        //bfs
        vector<vector<int>> visited=mat;
        queue<pair<pair<int, int>, int>> q;
        for(int i=0; i<mat.size(); i++)
        {
            for(int j=0; j<mat[0].size(); j++)
            {
                if(mat[i][j]==0)
                {
                    visited[i][j]=1;
                    q.push({{i, j}, 0});
                }
                if(mat[i][j]==1)
                {
                    visited[i][j]=0;
                }
            }
        }  

        while(!q.empty())
        {
            auto it=q.front();
            q.pop();
            int x=it.first.first;
            int y=it.first.second;
            int steps=it.second;
            int x4[4]={-1, 0, +1, 0};
            int y4[4]={0, +1, 0, -1};

            for(int i=0; i<4; i++)
            {
                int xp=x+x4[i];
                int yp=y+y4[i];
                if(xp>=0 && xp<mat.size() && yp>=0 && yp<mat[0].size() && visited[xp][yp]==0)
                {
                    visited[xp][yp]=1;
                    mat[xp][yp]=1+steps;
                    q.push({{xp, yp}, mat[xp][yp]});
                }
            }
        }  
        return mat;
    }
};
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