<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 Solution {
public:
    void addNeighbor(priority_queue<vector<int>, vector<vector<int>>, greater<>>& pq, vector<vector<int>>& grid, int x, int y){
        if (x >= (int)grid.size() || x < 0 || y >= (int)grid[0].size() || y < 0){
            return;
        }

        if (grid[x][y] != -1){
            pq.push({grid[x][y], x , y});
        }

        return;
    }

    int swimInWater(vector<vector<int>>& grid) {
        priority_queue<vector<int>, vector<vector<int>>, greater<>> pq;
        int targetx = (int)grid.size();
        int targety = (int)grid[0].size();

        pq.push({grid[0][0], 0, 0});
        int time = 0;
        while (pq.size()){
            vector<int> nextSquare = pq.top();
            pq.pop();

            time = max(nextSquare[0], time);
            printf("(%d %d %d)", nextSquare[0], nextSquare[1], nextSquare[2]);
            if (nextSquare[1] == targetx - 1 && nextSquare[2] == targety - 1){
                break;
            }

            grid[nextSquare[1]][nextSquare[2]] = -1;
            addNeighbor(pq, grid, nextSquare[1] + 1, nextSquare[2]);
            addNeighbor(pq, grid, nextSquare[1] - 1, nextSquare[2]);
            addNeighbor(pq, grid, nextSquare[1], nextSquare[2] + 1);
            addNeighbor(pq, grid, nextSquare[1], nextSquare[2] - 1);
        }


        return time;
    }
};
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