<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) {
        queue<pair<int,int>> q;
        int m=mat.size(),n = mat[0].size();
        vector<vector<int>> ans(m,vector<int>(n,1e5));
        vector<vector<int>> vis(m,vector<int>(n,0));
        for (int i=0;i<m;i++){
            for (int j=0;j<n;j++){
                if (!mat[i][j]) {
                    q.push({i,j});
                    vis[i][j]=1;
                    ans[i][j]=0;
                }
            }
        }
        while (!q.empty()){
            int curx = q.front().first;
            int cury = q.front().second;
            q.pop();
            vector<int> X = {0,0,1,-1},Y={1,-1,0,0};
            for (int p =0 ;p<4;p++){
                int newx = curx + X[p],newy = cury  + Y[p];
                if (newx >=0 && newx < m && newy >=0 && newy<n){
                    if (!vis[newx][newy]){
                    vis[newx][newy]=1;
                    ans[newx][newy] = min(ans[newx][newy],ans[curx][cury]+1);
                    q.push({newx,newy});
                    }
                }    
            }
        }
        return ans;
    }
};
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