<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();
        if(n == 1) return grid[0][0];

        int ans = grid[n-1][n-1];

        vector<vector<bool>> v(n,vector<bool>(n,false));


        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> q;
        q.push({grid[0][0],0,0});
        v[0][0] = true;



        while(!q.empty()) {
            int level = q.top()[0];
            int x = q.top()[1];
            int y = q.top()[2];
                        q.pop();

            ans = max(ans, level);


            vector<vector<int>> dir = {{0,1},{1,0},{0,-1},{-1,0}};

            for(auto d:dir) {
                int xx = x + d[0];
                int yy = y + d[1];
                if(xx == n-1 && yy == n-1) return ans;

                if(xx < 0 || yy < 0 || xx == n || yy == n || v[xx][yy]) continue;

                q.push({grid[xx][yy],xx,yy});
                v[xx][yy] = 1;

            }
        }
        return ans;
        
    }
};
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