<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>> costMatrix(mat.size(), vector<int>(mat[0].size(), INT_MAX));

        for(int i = 0; i < mat.size(); i++) {
            for(int j = 0; j < mat[0].size(); j++) {
                if(mat[i][j] == 0 && costMatrix[i][j] > 0) {
                    traverse(mat, costMatrix, i, j, 0);
                    break;
                }
            }
        }

        return costMatrix;
    }

    void traverse(vector<vector<int>>& mat, vector<vector<int>>& costMatrix, int row, int column, int cost) {
        if(row < 0 || row >= mat.size() || column < 0 || column >= mat[0].size()) {
            return;
        }

        // We already found a better path before
        if(cost >= costMatrix[row][column]) {
            return;
        }

        if(mat[row][column] == 0) {
            cost = 0;
        }

        costMatrix[row][column] = cost;

        traverse(mat, costMatrix, row + 1, column, cost+1);
        traverse(mat, costMatrix, row - 1, column, cost+1);
        traverse(mat, costMatrix, row, column + 1, cost + 1);
        traverse(mat, costMatrix, row, column - 1, cost + 1);
    }
};
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