<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 n,m;
    vector<vector<int>>dir{{-1,0} , {0,1} , {1,0} , {0,-1}}; //{rowOffset , colOffset}
    int swimInWater(vector<vector<int>>& grid) {
         n = grid.size(); m = grid[0].size();

         int ans = INT_MAX;
         int start = 0;
         int end   = INT_MAX;
         while(start <= end){
            int mid = (0LL + start + end)/(1LL * 2);
            if(bfs(mid,grid) == true) {
                ans = mid;
                end = mid - 1;
            }
            else {
                start = mid + 1;
            }
         }
         return ans;
    }
    bool isValid(int row , int col){
        if(row >= 0 and row < n  and col >=0 and col < m)
        return true;

        return false;
    }
     bool bfs(int &mid , vector<vector<int>>&arr){
         vector<vector<bool>>vis(arr.size(),vector<bool>(arr[0].size(),false));
         queue<pair<int,int>>q;
         if(arr[0][0] <= mid) {
            q.push({0,0});
            vis[0][0] = true;
         }
         while(!q.empty()){
             auto p = q.front(); q.pop();
             int row = p.first;
             int col = p.second;

             if(row == arr.size()-1 and col == arr[0].size()-1) return true;

             for(int i = 0 ; i < dir.size() ; i++){
                   if(isValid(row+dir[i][0] , col+dir[i][1]) == true and vis[row+dir[i][0]][col+dir[i][1]] == false and arr[row+dir[i][0]][col+dir[i][1]] <= mid){
                        q.push({row+dir[i][0],col+dir[i][1]});
                        vis[row+dir[i][0]][col+dir[i][1]] = true;
                   }
             }         
         }
         return false;
    }
};
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