<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:
    int m;
    int n;
    vector<vector<int>> directions = {{0,1}, {0,-1}, {1,0}, {-1, 0}};
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        m = mat.size();
        n = mat[0].size();
        vector<vector<int>> matrix(m, vector<int>(n, 0));
        vector<vector<bool>> seen(m, vector<bool>(n, false));

        queue<vector<int>> q;

        for(int row =0; row<m; row++){
            for(int col =0; col<n; col++){
                matrix[row][col] = mat[row][col];
                if(matrix[row][col]==0){
                    q.push({row,col,0});
                    seen[row][col] = true;
                }
            }
        }

        while(!q.empty()){
            vector<int> curr = q.front();
            q.pop();
            int row = curr[0];
            int col = curr[1];
            int steps = curr[2];

            for(vector<int>& direction: directions){
                int nextRow = row + direction[0], nextCol = col + direction[1];
                if(valid(nextRow, nextCol) && !seen[nextRow][nextCol]){
                    seen[nextRow][nextCol] = true;
                    q.push({nextRow, nextCol, steps+1});
                    matrix[nextRow][nextCol] = steps+1;
                }
            }
        }
        return matrix;
    }
    bool valid(int row, int col){
        return 0<=row && row<m && 0<=col && col<n;
    }
};
61506