<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();
        map<int,vector<int>> skip;
        
        int ans = grid[0][0];
        dfs(0,0,grid,ans,skip,n);
        return ans;
    }

    bool dfs(int r, int c, vector<vector<int>>& grid, int& ans, map<int,vector<int>>& skip, int& n){
        // cout<<r<<" "<<c<<endl;
        if(r == n-1 and c == n-1){
            ans = max(ans, grid[r][c]);
            return true;
        }
        ans = max(ans,grid[r][c]);
        grid[r][c] = -1;
        
        if(check(r-1,c,grid,skip, n))
            skip.insert(pair<int,vector<int>>(grid[r-1][c],{r-1,c}));
        if(check(r+1,c,grid,skip, n))
            skip.insert(pair<int,vector<int>>(grid[r+1][c],{r+1,c}));
        if(check(r,c-1,grid,skip, n))
            skip.insert(pair<int,vector<int>>(grid[r][c-1],{r,c-1}));
        if(check(r,c+1,grid,skip, n))
            skip.insert(pair<int,vector<int>>(grid[r][c+1],{r,c+1}));
        
        r = skip.begin()->second[0];
        c = skip.begin()->second[1];
        skip.erase(skip.begin());
        if(dfs(r,c, grid,ans,skip, n))
            return true;
        return false;
    }

    bool check(int r, int c, vector<vector<int>>& grid, map<int,vector<int>>& skip, int& n){
        if(r<0 or c<0 or r==n or c==n or grid[r][c] == -1 or skip.find(grid[r][c])!=skip.end())
            return false;
        return true;
    }
};
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