<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();
    int m = grid[0].size();
    vector<vector<int>> vis(n, vector<int>(m, 0));
    vector<vector<int>> dist(n, vector<int>(m, 0));
    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)
            {
                q.push({{i, j}, 0});
                vis[i][j] = 1;
            }
            else
            {
                vis[i][j] = 0;
            }
        }
    }

    while (!q.empty())
    {
        int row = q.front().first.first;
        int col = q.front().first.second;
        int steps = q.front().second;
        q.pop();
        dist[row][col] = steps;
        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        for (auto dir : directions)
        {
            int nr = row + dir.first;
            int nc = col + dir.second;

            if (nr >= 0 && nc >= 0 && nr < n && nc < m && vis[nr][nc] == 0)
            {
                vis[nr][nc] = 1;
                q.push({{nr, nc}, steps + 1});
            }
        }
    }
    return dist;
}

};
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