<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 {
    void solve(int ni, int nj, map<pair<int, int>, int>& q, vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int dis = INT_MAX;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    int x = ni - i;
                    int y = nj - j;
                    int nx = abs(x);
                    int ny = abs(y);
                    int d = ny + nx;
                    dis = min(dis, d);
                   
                }
            }
        }
        
        q[{ni, nj}] = dis;
    }
public:
    int maxDistance(vector<vector<int>>& grid) {
        map<pair<int, int>, int> q;
        int n = grid.size();
        int m = grid[0].size();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {  // Change `n` to `m`
                if (grid[i][j] == 0) {
                    solve(i, j, q, grid);  // Pass `grid` as an argument
                }
            }
        }
        int d = -1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {  
                if (grid[i][j] == 0) {
                    d = max(d, q[{i, j}]); 

                }

            }
        }
        if(d==2147483647)
        {
            return -1;
        }
        return d;
    }
};

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