<p>You are given an <code>n x n</code> integer matrix <code>grid</code> where each value <code>grid[i][j]</code> represents the elevation at that point <code>(i, j)</code>.</p>

<p>The rain starts to fall. At time <code>t</code>, the depth of the water everywhere is <code>t</code>. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most <code>t</code>. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.</p>

<p>Return <em>the least time until you can reach the bottom right square </em><code>(n - 1, n - 1)</code><em> if you start at the top left square </em><code>(0, 0)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim1-grid.jpg" style="width: 164px; height: 165px;" />
<pre>
<strong>Input:</strong> grid = [[0,2],[1,3]]
<strong>Output:</strong> 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim2-grid-1.jpg" style="width: 404px; height: 405px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
<strong>Output:</strong> 16
<strong>Explanation:</strong> The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
</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 &lt;= 50</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;&nbsp;n<sup>2</sup></code></li>
	<li>Each value <code>grid[i][j]</code> is <strong>unique</strong>.</li>
</ul>


class Time {
         public:
    int x;
    int y;
    int time;
  
    Time(int t, int x, int y) {
        this->time = t;
        this->x = x;
        this->y = y;
    }
 
    struct Comparator {
        bool operator()(Time a, Time b) {
            return (a.time > b.time);
        }
    };
};
class Solution {
public:
    int swimInWater(vector<vector<int>>& grid) {
        int row = grid.size();
        int col = grid[0].size();
        vector<vector<int>> dirs = {{0,1}, {-1,0}, {1,0}, {0, -1}};
        vector<vector<bool>> visited(row, vector<bool>(col, false));
        visited[0][0] = true;
        int res = 0;
        priority_queue<Time, vector<Time>, Time::Comparator> pq;
        pq.push(Time(grid[0][0], 0, 0));
        while (!pq.empty()) {
            auto curr = pq.top(); pq.pop();
            res = max (res, curr.time);
            if (curr.x == row-1 && curr.y == col-1) return res;
            cout << "Hello!" << endl;
            for (auto dir : dirs) {
                int crow = curr.x + dir[0];
                int ccol = curr.y + dir[1];
                 cout << "Google!" << endl;
                if (crow >= 0 && ccol >=0 && crow < row && ccol < col&& !visited[crow][ccol] ) {
                    visited[crow][ccol] = 1;
                   
                    pq.push(Time(grid[crow][ccol], crow, ccol));
                }
            }
        }
        return -1;
    }
};
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