<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();
        set<pair<int,int>> visited;
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq;
        vector<int> starting={grid[0][0],0,0};
        pq.push(starting);
        vector<vector<int>> directions={{0,1},{0,-1},{1,0},{-1,0}};

        visited.insert({0,0});

        while(!pq.empty()){
            vector<int> vec=pq.top();
            pq.pop();
            int t=vec[0],r=vec[1],c=vec[2];
            visited.insert({r,c});

            if(r==n-1 && c==n-1){
                return t;
            }
            for(auto d:directions){
                int neiRow=r+d[0],neiCol=c+d[1];
                if(neiRow<0 || neiRow>=n || neiCol<0 || neiCol>=n || visited.find({neiRow,neiCol})!=visited.end()){
                    continue;
                }
                visited.insert({neiRow,neiCol});
                vector<int> tmp;
                tmp.push_back(max(t,grid[neiRow][neiCol]));
                tmp.push_back(neiRow);
                tmp.push_back(neiCol);
                pq.push(tmp);
            }
        }
        return 0;
    }
};
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