<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:
    void bfs(vector<vector<int>>& mat,set<pair<int,int>>&st){
        queue<pair<pair<int,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){
                    q.push({{i,j},0});
                    st.insert({i,j});
                }
            }
        }

        while(!q.empty()){
            auto [frontNode,lvl] = q.front();
            q.pop();
            int dx[] = { -1,0,1,0};
            int dy[] = {0,1,0,-1};
            // int level = frontNode.second;

            for(int i=0;i<4;i++){

                int newx = frontNode.first + dx[i];
                int newy = frontNode.second + dy[i];

                if(newx>=0 && newx<mat.size() && newy>=0 && newy<mat[0].size() && st.find({newx,newy})==st.end()){
                    mat[newx][newy] = lvl+1;
                    q.push({{newx,newy},lvl+1});
                    st.insert({newx,newy});


                }
            }
        }
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {

        set<pair<int,int>>st;
        bfs(mat,st);

        return mat;
        
    }
};
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