<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) {
        int n = grid.size();
        vector<vector<int>> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        
        // Min-heap to store the current (time, x, y)
        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;
        
        // Visited array to mark cells we've processed
        vector<vector<bool>> visited(n, vector<bool>(n, false));
        
        // Start at (0, 0)
        pq.push({grid[0][0], {0, 0}});
        visited[0][0] = true;
        
        // Dijkstra's algorithm to find the minimum time to reach (n-1, n-1)
        while (!pq.empty()) {
            auto [time, coord] = pq.top();
            int x = coord.first, y = coord.second;
            pq.pop();
            
            // If we reach the bottom-right corner, return the time
            if (x == n - 1 && y == n - 1) return time;
            
            // Explore the four directions
            for (auto dir : directions) {
                int nx = x + dir[0], ny = y + dir[1];
                
                // Check if the new position is within bounds and not visited
                if (nx >= 0 && nx < n && ny >= 0 && ny < n && !visited[nx][ny]) {
                    visited[nx][ny] = true;
                    // Push the maximum time required to reach the new cell
                    pq.push({max(time, grid[nx][ny]), {nx, ny}});
                }
            }
        }
        
        return -1; // This should never be reached
    }
};
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