<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:
    int maxDistance(vector<vector<int>>& grid) {
        int n = grid.size();
        queue<int> bfs;
        bool allOnes = true;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    bfs.push(i * n + j);
                    grid[i][j] = 0;
                    
                } else {
                    allOnes = false;
                    grid[i][j] = -1;
                }
            }
        }
        if (allOnes) return -1;
        int level = 1;
        while (!bfs.empty()) {
            int size = bfs.size();
            for (int i = 0; i < size; ++i) {
                int coord = bfs.front();
                bfs.pop();
                int x = coord / n;
                int y = coord % n;
                if (x > 0 && grid[x - 1][y] == -1) {
                    grid[x - 1][y] = level;
                    bfs.push((x - 1)*n + y);
                }
                if (y > 0 && grid[x][y - 1] == -1) {
                    grid[x][y - 1] = level;
                    bfs.push(x*n + y - 1); 
                }
                if (x < n - 1 && grid[x + 1][y] == -1) {
                    grid[x + 1][y] = level;
                    bfs.push((x + 1)*n + y); 
                }
                if (y < n - 1 && grid[x][y + 1] == -1) {
                    grid[x][y + 1] = level;
                    bfs.push(x*n + y + 1); 
                }
            }
            level++;
        }
        int maxDist = -1;
        
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                cout << grid[i][j];
                maxDist = max(maxDist, grid[i][j]);
            }
        }
        return maxDist;
    }
};
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