<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:
    vector<int> dr = {1, 0, -1, 0};
    vector<int> dc = {0, -1, 0, 1};

public:
    bool isValid(int row, int col, int n) {
        return (row >= 0 && row < n && col >= 0 && col < n);
    }

    int maxDistance(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> vis(n, vector<int>(n, 1e9));
        set<pair<int, pair<int, int>>> st;
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j) {
                if(grid[i][j] == 1) {
                    st.insert({0, {i, j}});
                    vis[i][j] = 0;
                }
            }
        }

        int ans = -1;
        while(st.size() != 0) {
            auto it = *(st.begin());
            int dist = it.first;
            int x = it.second.first;
            int y = it.second.second;
            st.erase(it);

            if(grid[x][y] == 0) {
                ans = max(ans, vis[x][y]);
            } 
            for(int i = 0; i < 4; ++i) {
                int newX = x + dr[i];
                int newY = y + dc[i];
                if(isValid(newX, newY, n) && grid[newX][newY] == 0 && vis[newX][newY] > 1 + dist) {
                    vis[newX][newY] = 1 + dist;
                    st.insert({1 + dist, {newX, newY}});
                }
            }
        }
        return ans;
    }
};
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