<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(int i, int j, vector<vector<int>>&mat, vector<vector<int>>&dist){
//         int n = mat.size();
//         int m = mat[0].size();
//         queue<pair<int,int>>q1;
//         q1.push({i,j});
//         while(q1.size() != 0){
//             pair<int,int>node = q1.front();
//             q1.pop();
//             int x = node.first;
//             int y = node.second;
//             int arr[4][2] = {{-1,0},{1,0},{0,1},{0,-1}};
//             for(int k = 0; k<4; k++){
//                 int dx = x + arr[k][0];
//                 int dy = y + arr[k][1];
//                 if(dx >= 0 && dx < n && dy >= 0 && dy < m && (mat[dx][dy] != 0) && (dist[dx][dy] > dist[x][y]+1)){
//                     dist[dx][dy] = 1 + dist[x][y];
//                     q1.push({dx,dy});
//                 }
//             }
//         }
//         return;
//     }
//     vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
//         int n = mat.size();
//         int m = mat[0].size();
//         vector<vector<int>>vis(n, vector<int>(m,0));
//         vector<vector<int>>dist(n,vector<int>(m,1e9));
//         for(int i = 0; i<n; i++){
//             for(int j = 0; j<m; j++){
//                 if(mat[i][j] == 0){
//                     dist[i][j] = 0;
//                     bfs(i,j,mat,dist);
//                 }
//             }
//         }
//         return dist;
//     }
// };


class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        int n = mat.size();
        int m = mat[0].size();
        vector<vector<int>>vis(n, vector<int>(m,0));
        vector<vector<int>>ans(n, vector<int>(m,1e9));
        queue<pair<int,pair<int,int>>>q1;
        for(int i = 0; i<n; i++){
            for(int j = 0; j<m; j++){
                if(mat[i][j] == 0){
                    q1.push({i,{j,0}});
                    ans[i][j] = 0;
                    vis[i][j] = 1;
                }
            }
        }
        while(q1.size() != 0){
            int x = q1.front().first;
            int y = q1.front().second.first;
            int dist = q1.front().second.second;
            q1.pop();
            vis[x][y] = 1;
            int arr[4][2] = {{0,-1},{1,0},{0,1},{-1,0}};
            for(int k = 0; k<4; k++){
                int dx = x + arr[k][0];
                int dy = y + arr[k][1];
                if(dx >= 0 && dx < n && dy >= 0 && dy < m && vis[dx][dy] == 0){
                    ans[dx][dy] = min(ans[dx][dy], dist+1);
                    q1.push({dx,{dy,dist+1}});
                }
            }
        }
        return ans;
    }
};
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