<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:
    struct Comparator {
        bool operator()(pair<pair<int,int>, int> a,pair<pair<int,int>, int> b) {
            return (a.second > b.second );
        }
    };
    int swimInWater(vector<vector<int>>& grid) {

        set<pair<int,int>> visited;
        priority_queue<pair<pair<int,int>, int>, vector<pair<pair<int,int>, int>>, Comparator> q;

        q.push({{0,0},grid[0][0]});
        int ans = INT_MAX;
        //visited.insert({0,0});
        
        while(q.size() )
        {
            int r = q.top().first.first;
            int c = q.top().first.second;
            visited.insert({r,c});
            int value = q.top().second;
            cout << r << " " << c << " "<< value <<endl;
            
            q.pop();

            if(r == grid.size()-1 && c == grid[0].size()-1)
            {
                ans = min(ans, value);
            }

            if(r-1 >= 0 && visited.count({r-1,c})==0)
            {
                q.push({{r-1, c},max(value, grid[r-1][c])});
               
            }
            if(r+1 < grid.size() && visited.count({r+1,c})==0)
            {
                q.push({{r+1, c},max(value, grid[r+1][c])});
                 
            }
            if(c-1 >= 0 && visited.count({r,c-1})==0)
            {
                q.push({{r, c-1},max(value, grid[r][c-1])});
                 
            }
            if(c+1 < grid[0].size() && visited.count({r,c+1}) == 0)
            {
                q.push({{r, c+1},max(value, grid[r][c+1])});
            }
        }
        return ans;
        
    }
};


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