<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>>& mat) {
        int n = mat.size(),  m = mat[0].size(), ans = -1;
        vector<int> dirX = {0, 0 , 1, -1};
        vector<int> dirY = {1, -1 , 0, 0};

        queue<pair<int, int >> q;

        for(int i=0; i<n; i++){
                for(int j=0; j<m; j++){
                    if(mat[i][j]==1){
                        q.push({i,j});
                    }
                }
        }
        if (q.empty() || q.size() == n * n) 
            return -1;
        vector<vector<int>> vis(n, vector<int>(m, 0));
        while(!q.empty()){
            int size = q.size();
            while(size--){
                auto [x, y] = q.front();
                q.pop();
                for(int it = 0; it<4; it++){
                    int newX = x+dirX[it];
                    int newY = y+dirY[it];
                    if(newX >=0 && newX <n && newY >=0 && newY <m and vis[newX][newY]==0){
                        vis[newX][newY] =1;
                        q.push({newX, newY});
                    }
                }
            }
            ans++;
        }

        return ans; 

        
    }
};
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