<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();
     vector<vector<int>>ans(n,vector<int>(m,0));

     queue<pair<int,pair<int,int>>> q;
     vector<vector<bool>> vis (m,vector<bool>(n,false));

     for(int i =0;i<m;i++)
     {
        for(int j=0;j<n;j++)
        {
            if(mat[i][j]==0)
            {
                q.push({0,{i,j}});
                vis[i][j]=1;
            }
            //else vis[i][j] =0;
        }
     }

     
     while(!q.empty())
     {
        int dist = q.front().first;
        int x = q.front().second.first;
        int y = q.front().second.second;
        q.pop();

        vector<int> dirx = {-1,0,1,0};
     vector<int> diry = {0,-1,0,1};


        for(int i =0;i<4;i++)
        {
            int nx = x+dirx[i]; int ny = y+diry[i];

            if(nx>=0&&ny>=0 &&nx<m &&ny<n && !vis[nx][ny])
           { if(!vis[nx][ny]){
                    vis[nx][ny] =1;
                    mat[nx][ny]=dist+1;
                    q.push({dist+1,{nx,ny}});
                }}
        }
     }

     return mat;   
    }
};
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