<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 count = 0, step = 0;
        vector<vector<int>> ans = mat;
        vector<vector<bool>> vis = vector(mat.size(), vector(mat[0].size(), false));
        queue<pair<int, int>> q;
        for (int i = 0; i < mat.size(); i++) {
            for (int j = 0; j < mat[0].size(); j++) {
                if (mat[i][j] == 0) {
                    vis[i][j] = true;
                    q.push({i, j});
                    count++;
                }
            }
        }
        while(!q.empty() && count < mat.size() * mat[0].size()) {
            queue<pair<int, int>> q1;
            step++;
            while(!q.empty() && count < mat.size() * mat[0].size()) {
                int i = q.front().first, j = q.front().second;
                q.pop();
                if (i - 1 >= 0 && !vis[i - 1][j]) {
                    vis[i - 1][j] = true;
                    ans[i - 1][j] = step;
                    q1.push({i - 1, j});
                    count++;
                }
                if (i + 1 < mat.size() && !vis[i + 1][j]) {
                    vis[i + 1][j] = true;
                    ans[i + 1][j] = step;
                    q1.push({i + 1, j});
                    count++;
                }
                if (j - 1 >= 0 && !vis[i][j - 1]) {
                    vis[i][j - 1] = true;
                    ans[i][j - 1] = step;
                    q1.push({i, j - 1});
                    count++;
                }
                if (j + 1 < mat[0].size() && !vis[i][j + 1]) {
                    vis[i][j + 1] = true;
                    ans[i][j + 1] = step;
                    q1.push({i, j + 1});
                    count++;
                }
            }
            q = q1;
        }
        return ans;
    }
};
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