<p>Given an <code>n x n</code> <code>grid</code>&nbsp;containing only values <code>0</code> and <code>1</code>, where&nbsp;<code>0</code> represents water&nbsp;and <code>1</code> represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance.&nbsp;If no land or water exists in the grid, return <code>-1</code>.</p>

<p>The distance used in this problem is the Manhattan distance:&nbsp;the distance between two cells <code>(x0, y0)</code> and <code>(x1, y1)</code> is <code>|x0 - x1| + |y0 - y1|</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/1336_ex1.JPG" style="width: 185px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,1],[0,0,0],[1,0,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The cell (1, 1) is as far as possible from all the land with distance 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/1336_ex2.JPG" style="width: 184px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The cell (2, 2) is as far as possible from all the land with distance 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n&nbsp;&lt;= 100</code></li>
	<li><code>grid[i][j]</code>&nbsp;is <code>0</code> or <code>1</code></li>
</ul>

class Solution {
private: 
    pair<int, int> delimit = {-1, -1};
public:
    int maxDistance(vector<vector<int>>& grid) {
        vector<vector<int>> ans(grid.size(), vector<int>(grid[0].size(), INT_MIN));
        queue<pair<int, int>> next;

        int answer = -1;
        // Init queue

        for(int i=0;i<grid.size();i++) {
            for(int j=0;j<grid[i].size();j++) {
                if(grid[i][j] == 1) {
                    next.push({i, j});
                }
            }
        }

        next.push(delimit);

        int level = 0;
        while(!next.empty()) {
            pair<int, int> current = next.front();
            int i = current.first;
            int j = current.second;
            next.pop();
            if(current == delimit) {
                if(!next.empty()) {
                    next.push(delimit);
                    level++;
                }
            } else if(ans[i][j] == INT_MIN){
                ans[i][j] = level;
                if(i>0 && grid[i-1][j] == 0 && ans[i-1][j] == INT_MIN) {
                    next.push({i-1, j});
                }

                // Down 
                if(i<grid.size() - 1 && grid[i+1][j] == 0 && ans[i+1][j] == INT_MIN) {
                    next.push({i+1, j});
                }

                // Left
                if(j>0 && grid[i][j-1] == 0 && ans[i][j-1] == INT_MIN) {
                    next.push({i, j-1});
                }

                // Right
                if(j < grid[0].size() - 1 && grid[i][j+1] == 0 && ans[i][j+1] == INT_MIN) {
                    next.push({i, j+1});
                }
            }
        }

        for(int i=0;i<ans.size();i++) {
            for(int j=0;j<ans[i].size();j++) {
                if(grid[i][j] == 0 && ans[i][j] != INT_MIN) {
                    answer = max(ans[i][j], answer);
                }
            }
        }

        return answer;
    }
};
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