<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 m = mat.size(), n = mat[0].size();
        vector<vector<int>> dis(m, vector<int>(n, 0)), vis(m, vector<int>(n, -1));
        queue<pair<int, pair<int, int>>> q;
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(mat[i][j] == 0) {
                    vis[i][j] = 0;
                    q.push({0, {i,j}});
                }
            }
        }
        
        while(!q.empty()) {
            auto it = q.front();
            q.pop();
            int i = it.second.first, j = it.second.second;
            int steps = it.first;
            dis[i][j] = steps;
            vector<int> row = {-1,0,1,0}, col = {0,1,0,-1};
            for(int x=0; x<4; x++) {
                int ni = i+row[x], nj = j+col[x];
                if(ni>=0 && ni<m && nj>=0 && nj < n && vis[ni][nj] == -1 && mat[ni][nj] == 1) {
                    vis[ni][nj] = 0;
                    q.push({steps+1, {ni, nj}});
                }
            }
        }
        
        return dis;
    }
};
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