<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:
    vector<pair<int, int>> d = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    void dfs(pair<int, int> node, int& t, set<pair<int, int>>& visited, vector<vector<int>>& grid){
        visited.insert({node.first, node.second});

        for (auto [dx, dy] : d){
            pair<int, int> newNode = {node.first + dx, node.second + dy};

            if (visited.find({newNode.first, newNode.second}) != visited.end()) continue;
            if (newNode.first < 0 || newNode.first >= grid.size() || newNode.second < 0 || newNode.second >= grid.size()) continue;
            if (grid[newNode.first][newNode.second] > t) continue;

            dfs(newNode, t, visited, grid);
        }
    }

    int swimInWater(vector<vector<int>>& grid) {
        // DFS, con cada numero, si hay una distancia hasta (n-1, n-1) que no sea 1e9, retornar ese numero
        // Eso es O(n² * n²) = O(n⁴), sí pasa ya que n es pequeño pero se puede hacer búsqueda binaria para
        // reducirlo a O(n²log(n²))

        int n = grid.size();
        int lo = max(grid[0][0] - 1, grid[n-1][n-1] - 1), hi = n * n - 1, mid;

        if (n == 1) return grid[0][0];

        while (hi - lo > 1){
            mid = (lo + hi) / 2;

            set<pair<int, int>> visited;

            dfs({0, 0}, mid, visited, grid);

            if (visited.find({n - 1, n - 1}) == visited.end()){
                lo = mid;
            }
            else hi = mid;
        }

        return lo + 1;
    }
};
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