<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 ans;

    int multisourcebfs(vector<vector<int>> &grid, queue<pair<int, int>> &q) {
        queue<pair<int, int>> nq;

        while(!q.empty()) {
            auto [i, j] = q.front();
            q.pop();

            int arr[4][2] = {{1,0}, {0,1}, {-1,0},{0,-1}};

            for(int k = 0; k < 4; k++) {
                int row = i + arr[k][0], col = j + arr[k][1];

                if(row >= 0 && col >= 0 && row < grid.size() && col < grid.size() && grid[row][col] == 0){
                    nq.push({row, col});
                    grid[row][col] = 1;
                    }
            }
        }

        if(nq.empty())
            return 0;
        int hold = 1 + multisourcebfs(grid, nq);
        ans = max(hold, ans);
        return hold;
    }

    int maxDistance(vector<vector<int>>& grid) {
         queue<pair<int, int>> q;
         int fl = 0;
         for(int i = 0; i < grid.size(); i++) {
            for(int j = 0; j < grid[0].size(); j++) {
                if(grid[i][j] == 1)
                    q.push({i, j});
                if(grid[i][j] == 0)
                    fl = 1;
            }
        }

        if(!fl || q.empty())
            return -1;
        ans = INT_MIN;
        
        multisourcebfs(grid, q);
        return ans;
    }
};
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