<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:
    queue<pair<pair<int,int>,int>> points;
    vector<vector<int>> matrix;
    int rows=0;
    int cols=0;

    void processPoint(pair<pair<int,int>,int> point){
        int row = point.first.first;
        int col = point.first.second;
        int distance = point.second;
        if(matrix[row][col]==-1)
            matrix[row][col]=distance;
        if(row-1>=0){
            if(matrix[row-1][col]==-1){
                points.push({{row-1,col},distance+1});
            }
        }
        if(row+1<rows){
            if(matrix[row+1][col]==-1){
                points.push({{row+1,col},distance+1});
            }
        }
        if(col-1>=0){
            if(matrix[row][col-1]==-1){
                points.push({{row,col-1},distance+1});
            }
        }
        if(col+1<cols){
            if(matrix[row][col+1]==-1){
                points.push({{row,col+1},distance+1});
            }
        }

    }

    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        rows =mat.size();
        cols =mat[0].size();
        matrix = mat;
        for(int i=0;i<rows;i++){
            for(int j=0;j<cols;j++){
                if(matrix[i][j]==0){
                    points.push({{i,j},0});
                }else{
                    matrix[i][j]=-1;
                }
            }
        }
        while(!points.empty()){
            processPoint(points.front());
            points.pop();
        }
        return matrix;
    }
};
42966