<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();
        queue<pair<int,int>> q;
        vector<vector<int>> vis(n, vector<int> (m,0));
        for(int i= 0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mat[i][j]==0){
                    q.push({i,j});
                    vis[i][j] = 1;
                }
            }
        }
        int count = 0;
        int x[4] = {0,0,1,-1};
        int y[4] = {1,-1,0,0};
        while(!q.empty()){
            stack<pair<int,int>> s;
            while(!q.empty()){
                s.push({q.front().first,q.front().second});
                q.pop();
            }
            count++;
            while(!s.empty()){
                int i = s.top().first;
                int j = s.top().second;
                s.pop();
                for(int a=0;a<4;a++){
                    if(i+x[a]>=0 && i+x[a]<n && j+y[a]>=0 && j+y[a]<m && mat[i+x[a]][j+y[a]]==1){
                        if(vis[i+x[a]][j+y[a]]==0){
                            mat[i+x[a]][j+y[a]]=count;
                            vis[i+x[a]][j+y[a]]=1;
                            q.push({i+x[a],j+y[a]});
                        }
                    }
                }
            }
        }
        return mat;
    }
};
47601