<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:

    // bool fun(vector<vector<int> >&grid,int mid){
        
    //     int n=grid.size();
    //     int m = grid[0].size();
    //     vector<vector<int> >vis(n,vector<int>(m,0));

    //     if(grid[0][0] < mid)

    // }


    int swimInWater(vector<vector<int>>& grid) {
        int l=0,r=0;
        int n = grid.size();
        int m = grid[0].size();
        // for(int i=0;i<n;i++){
        //     for(int j=0;j<m;j++){
        //         r=max(r,grid[i][j]);
        //     }
        // }
        // int ans=0;
        // while(l<=r){
        //     int mid = (l+r)/2;
        //     if(possible(grid,mid)){
        //         ans = mid;
        //         r = mid-1;
        //     }else{
        //         l=mid+1;
        //     }
        // }
        // return ans;


        set<pair<int,pair<int,int> > >st;
        st.insert({grid[0][0],{0,0}});
        vector<vector<int> >dis(n,vector<int>(m,INT_MAX));
        dis[0][0] = grid[0][0];

        int dx[] = {1,0,-1,0};
        int dy[] = {0,1,0,-1};

        while(st.size()){
            auto it = *st.begin();
            int cur_height = it.first;
            int cur_x = it.second.first;
            int cur_y = it.second.second;

            st.erase(it);

            for(int i=0;i<4;i++){
                int nx = cur_x + dx[i];
                int ny = cur_y + dy[i];

                if(nx<0 || nx>=n || ny<0 || ny>=m || dis[nx][ny] <= cur_height){
                    continue;
                }

                st.erase({dis[nx][ny],{nx,ny}});
                dis[nx][ny] = max(cur_height,grid[nx][ny]);
                st.insert({dis[nx][ny],{nx,ny}});
            }

        }
        return dis[n-1][m-1];

    }
};
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