<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>

#define ll int 
class Solution {
public:
   bool ok(ll i , ll j , ll &n, ll &m){
       if(i>=0&&i<n&&j>=0&&j<m){
        return true;
       }
       return false;
   }
    int swimInWater(vector<vector<int>>& grid) {
         vector<vector<int>>v = grid;
         ll n = v.size();
         ll m = v[0].size();
         priority_queue<vector<ll>,vector<vector<ll>>,greater<vector<ll>>>q;
         q.push({v[0][0],0,0});
          vector<pair<ll,ll> >p= {{0,-1},{0,1},{1,0},{-1,0}};
          vector<vector<ll> >vis(n+1,vector<ll>(m+1,1e9));
          vis[0][0]=v[0][0];
         while(!q.empty()){
             auto at = q.top();
             ll val = at[0];
             ll i = at[1];
             ll j = at[2];
             q.pop();
             for(auto bt:p){
                ll x = i+bt.first;
                ll y = j+bt.second;
                if(ok(x,y,n,m)){
                    ll diff = v[x][y]-val;
                    if(val>=v[x][y]&&vis[x][y]>val){
                         vis[x][y]= val;
                         q.push({val,x,y});
                    }
                    else if(val<v[x][y]&&vis[x][y]>val+diff){
                          vis[x][y]= val+diff;
                          q.push({val+diff,x,y}); 
                    }
                }
             }
            
         }
         return vis[n-1][m-1];

    }
};
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