<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 iSize = grid.size();
        vector<int> vCost(iSize*iSize, INT32_MAX);
        unordered_map<int, vector<pair<int, int>>> vMap;
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pqNodes;

        int iNode = 0;

        for(int r=0; r<iSize; r++)
        {
            for(int c=0; c<iSize; c++)
            {
                if(r-1>=0)
                    vMap[iNode].push_back({grid[r-1][c], iNode-iSize});
                if(r+1<iSize)
                    vMap[iNode].push_back({grid[r+1][c], iNode+iSize});
                if(c-1>=0)
                    vMap[iNode].push_back({grid[r][c-1], iNode-1});
                if(c+1<iSize)
                    vMap[iNode].push_back({grid[r][c+1], iNode+1});

                iNode++;
            }
        }

        vCost[0] = grid[0][0];

        pqNodes.push({grid[0][0],0});

        while(!pqNodes.empty())
        {
            int iCost = pqNodes.top().first, iNode = pqNodes.top().second;
            pqNodes.pop();
            
            for(auto it: vMap[iNode])
            {
                int iMax = max(iCost, it.first);
                if(vCost[it.second] > iMax)
                {
                    vCost[it.second] = iMax;
                    pqNodes.push({iMax, it.second});
                }
            }
        }
        return vCost[(iSize*iSize)-1];
    }
};
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