<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) {
        vector<vector<int>> res;
        queue<vector<int>> discovery_nodes;
        vector<vector<int>> dirs = {{0,1}, {0,-1}, {1,0}, {-1,0}};

        // init res
        for(int i=0; i<mat.size(); i++){
            vector<int> col;
            for(int j=0; j<mat[0].size(); j++){
                if(mat[i][j] == 0){discovery_nodes.push({i, j}); col.push_back(0);}
                else col.push_back(-1);
            }
            res.push_back(col);
        }

        // discover using breadth first
        while(!discovery_nodes.empty()){
            int i=discovery_nodes.front()[0], j=discovery_nodes.front()[1];
            discovery_nodes.pop();
            for(int d=0; d<dirs.size(); d++){
                //check boundaries
                if(i+dirs[d][0] < 0 || j+dirs[d][1] < 0 || i+dirs[d][0] > res.size()-1 || j+dirs[d][1] > res[0].size()-1) continue;
                int dist = res[i][j] + 1;
                if(res[i+dirs[d][0]][j+dirs[d][1]] == -1){
                    res[i+dirs[d][0]][j+dirs[d][1]] = dist;
                    discovery_nodes.push({i+dirs[d][0], j+dirs[d][1]});
                }
                else res[i+dirs[d][0]][j+dirs[d][1]] = min(res[i+dirs[d][0]][j+dirs[d][1]], dist);
            }
        }

        return res;
    }
};
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