<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) {
        vector<vector<int>> ans(mat.size());
        vector<pair<int,int>>vec;
        for(int i = 0 ;i < mat.size();++i){
            for(int j = 0; j < mat[i].size();++j){
                ans[i].push_back(-1);
                if(mat[i][j] == 0){
                    vec.push_back(make_pair(i,j));
                }
            }
        }
        int value = 0;
        while(!vec.empty()){
            int sz = vec.size();
            for(pair<int,int> node:vec){
                ans[node.first][node.second] = value;
            }
            vector<pair<int,int>>temp;
            for(pair<int,int> node:vec){
                ans[node.first][node.second] = value;
                if(node.first - 1 >= 0 && ans[node.first-1][node.second] == -1){
                    temp.push_back(make_pair(node.first-1,node.second));
                }
                if(node.first + 1 < mat.size() && ans[node.first+1][node.second] == -1){
                    temp.push_back(make_pair(node.first+1,node.second));
                }
                if(node.second - 1 >= 0 && ans[node.first][node.second-1] == -1){
                    temp.push_back(make_pair(node.first,node.second-1));
                }
                if(node.second + 1 < mat[node.first].size() && ans[node.first][node.second+1] == -1){
                    temp.push_back(make_pair(node.first,node.second+1));
                }
            }
             vec.clear();
                vec = temp;
                temp.clear();
            value++;
        }
        return ans;
    }
};
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