<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 {
public:
    static inline int MAXI = INT_MAX/2;
    static inline int MINI = INT_MIN/2;
    static inline vector<int> dx = {1, 0, -1, 0};
    static inline vector<int> dy = {0, 1, 0, -1};
    
    bool isValid(int x, int y, int m, int n) {
        return x>=0 && y>=0 && x<m && y<n;
    }
    
    int maxDistance(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        
        vector<vector<bool>> vis(m, vector<bool>(n, false));
        vector<vector<int>> dis(m, vector<int>(n, MAXI));
        queue<pair<int, int>> q;
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(grid[i][j]) {
                    q.push({i, j});
                    dis[i][j] = 0;
                    vis[i][j] = true;
                }
            }
        }
        
        while(!q.empty()) {
            pair<int, int> p = q.front();
            q.pop();
            
            int x = p.first, y = p.second;
            for(int i=0; i<4; i++) {
                int u = x + dx[i], v = y + dy[i];
                
                if(!isValid(u, v, m, n) || vis[u][v]) {
                    continue;
                }
                
                dis[u][v] = min(dis[u][v], 1 + dis[x][y]);
                vis[u][v] = true;
                q.push({u, v});
            }
        }
        
        int maxi = MINI;
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(!grid[i][j]) {
                    maxi = max(maxi, dis[i][j]);
                }
            }
        }
        
        if(maxi == MINI || maxi == MAXI) {
            return -1;
        }
        
        return maxi;
    }
};
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