<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:

    bool validPoint(int p, int q, int n, int m){
        if(p<0 || p>=n || q<0 || q>=m) return false;
        return true;
    }

    int maxDistance(vector<vector<int>>& grid) {
        
        int n = grid.size(), m = grid[0].size();
        queue<pair<pair<int,int>, int>> q;
        vector<vector<int>> ans(n, vector<int>(m, INT_MAX));

        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(grid[i][j]==1){
                    q.push({{i,j}, 0});
                    ans[i][j] = 0;
                } 
            }
        }

        vector<int> ti = {1, -1, 0, 0, 1, 1, -1, -1};
        vector<int> tj = {0, 0, 1, -1, 1, -1, 1, -1};

        while(!q.empty()){
            pair<int,int> ind = q.front().first;
            int dis = q.front().second;
            q.pop();
            int ci = ind.first;
            int cj = ind.second;

            for(int k = 0; k < 8; k++){
                int x = ci+ti[k];
                int y = cj+tj[k];
                int newDist = dis + abs(x-ci) + abs(y-cj);
                if(validPoint(x, y, n, m) && grid[x][y] == 0 && ans[x][y] > newDist) {
                    ans[x][y] = newDist;
                    q.push({{x,y}, newDist});
                }
            }
        }

        int maxi = -1;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                maxi = max(maxi, ans[i][j]);
            }
        }

        return maxi == 0 || maxi == INT_MAX ? -1 : maxi;
    }
};
29918