<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 MOD 1000000007
#define pb push_back
#define _pb pop_back
#define vpii vector<pair<int, int>>
#define vi vector<int>
#define vvi vector<vector<int>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vs vector<string>
#define vvs vector<vector<string>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define ll long long
#define L left
#define R right
#define D data
#define V val
#define F first
#define S second
#define ii int, int
#define pii pair<int, int>
#define SIZE(arr) (sizeof(arr)/sizeof(arr[0]))
#define max3_(a, b, c) (max(a, max(b, c)))
#define min3_(a, b, c) (min(a, min(b, c)))
#define addMod(a, b) (int)( (ll)(a)%MOD + (b)%MOD )%MOD
#define prdMod(a, b) (int)( (ll)(a)%MOD * (b)%MOD )%MOD
#define repIt(it, v) for(auto it=v.begin(); it!=v.end(); it++)
#define rep(i, a, b) for(i=a; i<b; i++)
#define rep_(i, a, b) for(int i=a; i<b; i++)
#define repr(i, b, a) for(i=b; i>=a; i--)
#define repr_(i, b, a) for(int i=b; i>=a; i--)
#define umap unordered_map
#define MIN INT_MIN
#define MAX INT_MAX
#define INF 0x3f3f3f3f
#define umap unordered_map
typedef long long i64;
const double eps = 1e-9;
const double Pi = 2.0 * acos(0.0);

class Solution {
public:
    int swimInWater(vector<vector<int>>& mat) 
    {
        if( !mat.size() || !mat[0].size() ) return 0;

        int n=mat.size(), m=mat[0].size();
        int dx[4] = {-1, 0, 0, 1};
        int dy[4] = {0, -1, 1, 0};
        vvi dist(n, vi(m, INF));
        vvb visited(n, vb(m, 0));
        set<vi> st;
        int i, j, k, d, x, y, D;

        dist[0][0] = mat[0][0];
        st.insert({dist[0][0], 0, 0});

        while(!st.empty())
        {
            vi t = *st.begin();     st.erase(st.begin());
            i=t[1], j=t[2];

            if(visited[i][j]) continue;
            visited[i][j] = 1;

            if(i==n-1 && j==m-1) break;

            rep(k, 0, 4)
            {
                x= i+dx[k], y= j+dy[k];
                if(x>=0 && x<n && y>=0 && y<m
                    && !visited[x][y])
                {
                    D = max(dist[i][j], mat[x][y]);
                    if(dist[x][y]==INF || D<dist[x][y])
                    {
                        dist[x][y] = D;
                        st.insert({dist[x][y], x, y});
                    }
                }
            }

        }

        return dist[n-1][m-1];        
    }


};
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