<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 {

private:

    vector<vector<int>> dirs = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}};

    bool isValid(vector<vector<int>>& grid, int i, int j) {
        return i >= 0 && i < grid.size() && j >= 0 && j < grid[0].size();
    }

    int findLeastMaxPathValue(vector<vector<int>>& grid) {

        set<pair<int, pair<int, int>>> minSet;
        minSet.insert({grid[0][0], {0, 0}});

        vector<vector<int>> dist(grid.size(), vector<int>(grid[0].size(), INT_MAX));

        dist[0][0] = grid[0][0];

        while (!minSet.empty()) {

            auto minDistPair = *minSet.begin();
            minSet.erase(minSet.begin());

            auto [i, j] = minDistPair.second;

            for (auto dir: dirs) {
                if (isValid(grid, i+dir[0], j+dir[1]) && 
                    max(dist[i][j], grid[i+dir[0]][j+dir[1]]) < dist[i+dir[0]][j+dir[1]]) {
                    auto it = minSet.find({dist[i+dir[0]][j+dir[1]], {i+dir[0], j+dir[1]}});
                    if (it != minSet.end()) {
                        minSet.erase(it);
                    }
                    dist[i+dir[0]][j+dir[1]] = max(dist[i][j], grid[i+dir[0]][j+dir[1]]);
                    minSet.insert({dist[i+dir[0]][j+dir[1]], {i+dir[0], j+dir[1]}});
                }
            }

        }

        return dist[grid.size()-1][grid[0].size()-1];

    }

public:
    
    int swimInWater(vector<vector<int>>& grid) {
        return findLeastMaxPathValue(grid);
    }
};
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