<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();
        vector<vector<int>> vis(n ,vector<int> (n, 0));
        vector<vector<int>> dis(n, vector<int>(n));
        queue<pair<int, pair<int, int>>>q;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(grid[i][j] == 1)
                {
                    q.push({0, {i, j}});
                    vis[i][j] = 1;
                    dis[i][j] = 0;
                }
            }
        }
        int dr[] = {0, -1, 0, 1};
        int dc[] = {1, 0, -1, 0};
        int ans = 0;
        while(!q.empty())
        {
            int r = q.front().second.first;
            int c = q.front().second.second;
            int d = q.front().first;
            q.pop();
            ans = max(ans, d);
            for(int i = 0; i < 4; i++)
            {
                int nr = r + dr[i];
                int nc = c + dc[i];
                if(nr >= 0 && nc >= 0 && nr < n && nc < n)
                {
                    if(!vis[nr][nc])
                    {
                        dis[nr][nc] = 1 + d;
                        vis[nr][nc] = 1;
                        q.push({1 + d, {nr, nc}});
                    }
                    else
                    {
                        dis[nr][nc] = min(dis[nr][nc], 1 + d);
                    }
                }
            }
        }
        if(ans == 0) return -1;
        return ans;
    }
};









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