<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 t = 0;
        int m = grid.size();
        int n = grid[0].size();
        int dirs[5] = {0, 1, 0, -1, 0};
        if(m == 1 && n == 1) return grid[0][0];
        priority_queue<int, vector<int>, greater<int>> pq;
        unordered_map<int, pair<int, int>> mp;
        unordered_set<string> visited;
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                mp[grid[i][j]] = {i, j};
            }
        }
        t = grid[0][0];
        pq.push(grid[0][0]);
        visited.insert("0_0");
        while(true) {
            int i = mp[pq.top()].first;
            int j = mp[pq.top()].second;
            pq.pop();
            if(grid[i][j] > t) t = grid[i][j];
            for(int dir=0; dir<4; dir++) {
                int i2 = i + dirs[dir];
                int j2 = j + dirs[dir+1];
                string pos = to_string(i2) + "_" + to_string(j2);
                if(i2<0 || i2 == m || j2<0 || j2 == n || visited.find(pos) != visited.end()) continue;
                if(i2 == m-1 && j2 == n-1) return max(t, grid[i2][j2]);
                visited.insert(pos);
                pq.push(grid[i2][j2]);
            }
        }
        return -1;
    }
};
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