<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>>& grid) {
        int n = grid.size(); int m = grid[0].size();


        vector<vector<int>> dist(n ,vector<int> (m, INT_MAX));
        vector<vector<int>> vis(n, vector<int> (m, 0));

        queue<pair<pair<int,int>,int>> q;
        
        for(int i = 0;i<n;i++){
            for(int j = 0;j<m;j++){
                if(grid[i][j]==0) {
                    q.push({{i, j}, 0});
                    vis[i][j] = 1;
                }
            }
        }

        vector<pair<int,int>> moves = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};

        while(!q.empty()){
            int i = q.front().first.first;
            int j = q.front().first.second;
            int d = q.front().second;
            q.pop();
            dist[i][j] = min(d, dist[i][j]);
            vis[i][j] = 1;
            //cout<<i<<" "<<j<<" "<<dist[i][j]<<endl;
            for(auto x : moves){
                int nx = i+x.first;
                int ny = j+x.second;
               
                if(nx>=0 && ny>=0 && nx<n && ny<m && grid[nx][ny]==1 && vis[nx][ny]==0){
                    //cout<<"as";
                    q.push({{nx, ny}, d+1});
                }
            }
        }

        return dist;

    
    }
};
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