<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 m=mat.size();
        int n=mat[0].size();
        queue<pair<int,int>> que;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(mat[i][j]==0)
                {
                    que.push({i,j});
                }
                else
                {
                    mat[i][j]=INT_MAX;
                }
            }
        }

        while(!que.empty())
        {
            int size=que.size();

            vector<pair<int,int>> direction={{0,-1},{0,1},{-1,0},{1,0}};
            for(int i=0;i<size;i++)
            {
                auto x=que.front();
                int row=x.first;
                int col=x.second;
                que.pop();
                for(int j=0;j<4;j++)
                {
                   int newrow=row+direction[j].first;
                   int newcol=col+direction[j].second;
                   if(newrow>=0 && newrow<m && newcol>=0 && newcol<n && mat[newrow][newcol] > mat[row][col]+1)
                   {
                        mat[newrow][newcol]=mat[row][col]+1;
                        que.push({newrow,newcol});
                   }
                }
            }
        }
        return mat;
    }
};
34469