<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>

#include <bits/stdc++.h>
using namespace std;

class Solution {
    int direc[4][2] = {
        {0, 1},
        {0, -1},
        {1, 0},
        {-1, 0}
    };

    bool pathExists(vector<vector<int>>& grid, int time) {
        int n = grid.size();
        if (grid[0][0] > time) return false;
        
        queue<int> bfs;
        bfs.push(0);
        unordered_set<int> visited;
        visited.insert(0);
        
        while (!bfs.empty()) {
            int curr = bfs.front();
            bfs.pop();
            
            if (curr == n * n - 1) return true;
            
            int i = curr / n;
            int j = curr % n;

            for (int k = 0; k < 4; k++) {
                int ni = i + direc[k][0];
                int nj = j + direc[k][1];
                
                int next = ni * n + nj;
                if (ni >= 0 && ni < n && nj >= 0 && nj < n && grid[ni][nj] <= time && visited.count(next) == 0) {
                    bfs.push(next);
                    visited.insert(next);
                }
            }
        }
        return false;
    }

public:
    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        int start = grid[0][0];
        int end = n * n - 1;
        int res = end;

        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (pathExists(grid, mid)) {
                res = mid;
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }

        return res;
    }
};

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