<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) {
        int rows = mat.size();
        int cols = mat[0].size();
        
        queue<pair<pair<int,int>,int>> q; // Queue to store {i,j,distance}
        
        // Initialize the queue with all zero cells
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (mat[i][j] == 0) {
                    q.push({{i, j}, 0});
                }
            }
        }
        
        // Initialize the distance matrix with INT_MAX
        vector<vector<int>> dist(rows, vector<int>(cols, INT_MAX));
        
        while (!q.empty()) {
            auto p = q.front();
            q.pop();
            int i = p.first.first;
            int j = p.first.second;
            int d = p.second;
            
            // If the current cell is unvisited (still at INT_MAX), update its distance
            if (dist[i][j] == INT_MAX) {
                dist[i][j] = d;
                
                // Explore the neighboring cells and push them into the queue
                if (i > 0) {
                    q.push({{i - 1, j}, d + 1});
                }
                if (j > 0) {
                    q.push({{i, j - 1}, d + 1});
                }
                if (i < rows - 1) {
                    q.push({{i + 1, j}, d + 1});
                }
                if (j < cols - 1) {
                    q.push({{i, j + 1}, d + 1});
                }
            }
        }
        
        return dist;
    }
};

47601