<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>>& grid) {
        int n = grid.size(), m = grid[0].size(); 
        vector<vector<bool>> visited(n, vector<bool>(m, false));
        queue<pair<pair<int, int>, int>> q;

        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == 0) {
                    visited[i][j] = 1;
                    q.push({{i, j}, 0});
                }
            }
        }

        while(!q.empty()) {
            int currX = q.front().first.first;
            int currY = q.front().first.second;
            int dist = q.front().second;
            q.pop();

            vector<int> x = {1, -1, 0, 0};
            vector<int> y = {0, 0, 1, -1};

            for(int i = 0; i < 4; i++) {
                int newX = currX + x[i], newY = currY + y[i];

                if(newX < 0 || newX >= n || newY < 0 || newY >= m) continue;

                if(visited[newX][newY] == 0) {
                    grid[newX][newY] = dist + 1;

                    visited[newX][newY] = 1;
                    q.push({{newX, newY}, dist + 1});
                }
            }
        }

        return grid;
    }
};
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