<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 n = mat.size();
        int m = mat[0].size();
        vector<vector<int>>ans(n, vector<int>(m,0));
        queue<pair<int,int>>q;
        for (int i = 0;i<n;i++) {
            for (int j = 0;j<m;j++){
                if(mat[i][j] == 0) {
                    q.push({i,j});
                }
                else {
                    mat[i][j] = -1;
                }
            }
        }
        int level = 0;
        vector<int>x_moves = {1,0,-1,0};
        vector<int>y_moves = {0,-1,0,1};
        int sz;
        while(!q.empty()) {
            sz = q.size();
            for (int i = 0;i<sz;i++) {
                auto it = q.front();
                q.pop();
                int x = it.first;
                int y = it.second;
                if (mat[x][y] == -1) {
                    mat[x][y] = level;
                }
                for (int j = 0;j<4;j++){
                    int newX = x + x_moves[j]; 
                    int newY = y + y_moves[j];
                    if (newX >= 0 && newY >= 0 && newX < n && newY < m && mat[newX][newY] == -1) {
                        q.push({newX, newY});
                    } 
                }
            }
            level++;
        }
        return mat;
    }
};
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