<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:
    vector<vector<int>> dirs = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
    unordered_map<int, pair<int, int>> at;
    unordered_map<int, int> leader;
    int m, n;
    int conv(int i, int j)
    {
        return i*n+j;
    }
    int find_leader(int a)
    {
        if(leader[a]==a)
        return leader[a];
        return leader[a]=find_leader(leader[a]);
    }
    void do_union(int a, int b)
    {
        int pa = find_leader(a);
        int pb = find_leader(b);

        if(pa==pb)
        return;
        leader[pb]=pa;
    }
    bool check(int i, int j)
    {
        if(i<0 || j<0 || i>=m || j>=n)
        return false;
        return true;
    }
    void connect(vector<vector<int>>& grid, int i, int j, int time)
    {
        int p = conv(i, j);
        for(auto dir:dirs)
        {
            int x = i+dir[0], y = j+dir[1];
            int q = conv(x, y);
            if(check(x, y) && time>=grid[x][y])
            {
                do_union(p, q);
            }
        }
    }
    int swimInWater(vector<vector<int>>& grid) {
        m = grid.size(), n = grid[0].size();

        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                int x = conv(i, j);
                at[grid[i][j]]={i, j};
                leader[x]=x;
            }
        }
        int start = conv(0, 0);
        int end = conv(m-1, n-1);

        for(int time=0;time<2500;time++)
        {
            int i = at[time].first;
            int j = at[time].second;
            connect(grid, i, j, time);
            int u = find_leader(start);
            int v = find_leader(end);

            if(u==v)
            return time;
        }
        return -1;
    }
};
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