<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 rows[4] = {1,-1,0,0};
    int cols[4] = {0,0,1,-1};

    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int m = mat.size();
        int n = mat[0].size();
        vector<vector<int>> ans(m, vector<int>(n,0));
        vector<vector<int>> visited(m, vector<int>(n,0));
        queue<int> q;

        for (int i=0; i<m; i++){
            for (int j=0; j<n; j++){
                if (mat[i][j] == 0){
                    q.push(i*n+j);
                }
            }
        }

        q.push(-1);
        int ct = 0;

        while (!q.empty()){
            int front = q.front();
            q.pop();
            if (front == -1){
                if (!q.empty()){
                    q.push(-1);
                }
                ct++;
                continue;
            }

            int r = front/n, c = front%n;
            if (visited[r][c] == 1){
                continue;
            }

            visited[r][c] = 1;
            ans[r][c] = ct;

            for (int i=0; i<4; i++){
                int newrow = rows[i]+r;
                int newcol = cols[i]+c;

                if (newrow >=0 && newcol >= 0 && newrow < m && newcol < n){
                    q.push(newrow*n+newcol);
                }
            }
        }

        return ans;
    }
};
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