<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:
	int swimInWater(vector<vector<int>>& grid) {
		const int row = grid.size();
		const int col = grid.front().size();
		priority_queue<pair<int, vector<int>>, vector<pair<int, vector<int>>>, greater<>> min_heap;

		vector<vector<bool>> visited(row);
		for (int i = 0; i < row; ++i) {
			visited[i] = vector<bool>(col, false);
		}

		min_heap.emplace(grid[0][0], vector {0, 0});
		int result = 0;

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

		int t = 0;
		while (!min_heap.empty()) {
			const int height = min_heap.top().first;
			const auto grid_pos = min_heap.top().second;
			const int r = grid_pos[0];
			const int c = grid_pos[1];
			if (t < height) {
				++t;
				continue;
			}
			min_heap.pop();
			if ( r == row -1 && c == col - 1 )
				break;
			if (visited[r][c])
				continue;

			visited[r][c] = true;

			for (const auto& direction : directions) {
				const int x = r + direction.first;
				const int y = c + direction.second;

				if (x < 0 || y < 0 || x >= row || y >= col || visited[x][y])
					continue;

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

		return t;
	}
};

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