<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 n = grid.size();
        int lo = 0;
        int hi = n*n;

        int ans = 0;
        while(lo <= hi) {
            int mid = (lo + hi)/2;
            if(ok(grid, mid)) {
                hi = mid - 1;
                ans = mid;
            }else {
                lo = mid + 1;
            }
        }
        
        return ans;
    }

    bool ok(vector<vector<int>>& grid, int val) {
        int sx = 0;
        int sy = 0;

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

        queue<pair<int, int>> q;
        int n = grid.size();
        vector<vector<bool>> visited(n , vector<bool>(n, false));

        if (grid[0][0] <= val) { 
            visited[0][0] = true;
            q.push({0, 0});
        }
        while(!q.empty()) {
            auto [cr, cc] = q.front();
            q.pop();

            for(auto [dr, dc] : dirs){
                int nr = cr + dr;
                int nc = cc + dc;
                if(nr == n - 1 && nc == n - 1 && grid[nr][nc] <= val) {
                    return true;
                };
                if(indexOk(nr, nc, n, n) && !visited[nr][nc] && grid[nr][nc] <= val) {
                    q.push({nr, nc});
                    visited[nr][nc] = true;
                }
            }
        }

        return false;
    }

    bool indexOk(int r, int c, int n, int m) {
        return (r >= 0) && (c >= 0) && (r < m) && (c < n);
    }
};
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