<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 maxheightinpath=grid[0][0];
        if(grid.size()==1){
            return maxheightinpath;
        }
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>> pq;

        int n = grid.size();
        pq.push(make_pair(grid[0][0],make_pair(0,0)));

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

        while(true){
            auto x = pq.top();
            pq.pop();
            int height = x.first;
            maxheightinpath=max(height,maxheightinpath);
            pair<int,int> index = x.second;
            grid[index.first][index.second] = INT_MAX;

            for(auto x : adjacents(index , grid.size())){
                auto temp = make_pair(grid[x.first][x.second],x);
                pq.push(temp);
                grid[x.first][x.second]=INT_MAX;
                if((x.first==n-1)&&(x.second==n-1)){
                    return max(maxheightinpath,finalheight);
                }
            }
        }

        return max(maxheightinpath,finalheight);

    }

    vector<pair<int,int>> adjacents(pair<int,int> tocheck,int n){
        vector<pair<int,int>> ans;


        if(tocheck.second+1<n){
            ans.push_back(make_pair(tocheck.first,tocheck.second+1));
        }

        if(tocheck.second>0){
            ans.push_back(make_pair(tocheck.first,tocheck.second-1));
        }
        if(tocheck.first+1<n){
            ans.push_back(make_pair(tocheck.first+1,tocheck.second));
        }
        if(tocheck.first>0){
            ans.push_back(make_pair(tocheck.first-1,tocheck.second));
        }


        return ans;
    }
};
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