<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 {
	const vector<vector<int>> directions {
		{ 0,	1},
		{ 0, -1},
		 { 1,	 0},
		 {-1,  0}
	};

  public:
	int swimInWater(vector<vector<int>>& grid) {
		const int row_size = grid.size();
		const int col_size = grid.front().size();
		if (row_size == 1)
			return 0;
		priority_queue<vector<int>, vector<vector<int>>, greater<>> min_heap;
		vector visited(row_size, vector(col_size, false));
		visited[0][0] = true;
		min_heap.emplace(vector {grid[0][0], 0, 0});

		int result = 0;
		while (!min_heap.empty()) {
			const vector<int> curr = min_heap.top();
			min_heap.pop();

			result = max(result, curr[0]);

			for (int i = 0; i < 4; ++i) {
				const int x = curr[1] + directions[i][0];
				const int y = curr[2] + directions[i][1];

				if (x < 0 || y < 0 || x >= row_size || y >= col_size || visited[x][y])
					continue;

				if (x == row_size - 1 && y == col_size - 1)
					return max(result, grid[x][y]);

				min_heap.emplace(vector {grid[x][y], x, y});
				visited[x][y] = true;
			}
		}

		return -1;
	}
};

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