<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:
    void dfs(vector<vector<int>>& grid, int i, int j, vector<vector<int>>& dist){
        int m = grid.size(), n = grid[0].size();
        if(i > 0 && dist[i-1][j] > dist[i][j]){
            dist[i-1][j] = max(dist[i][j], grid[i-1][j]);
            dfs(grid, i-1, j, dist);
        } 

        if(j > 0 && dist[i][j-1] > dist[i][j]){
            dist[i][j-1] = max(dist[i][j], grid[i][j-1]);
            dfs(grid, i, j-1, dist);
        }

        if(i < m-1 && dist[i+1][j] > dist[i][j]){
            dist[i+1][j] = max(dist[i][j], grid[i+1][j]);
            dfs(grid, i+1, j, dist);
        }

        if(j < n-1 && dist[i][j+1] > dist[i][j]){
            dist[i][j+1] = max(dist[i][j], grid[i][j+1]);
            dfs(grid, i, j+1, dist);
        }
    }

    int swimInWater(vector<vector<int>>& grid) {
        int N = grid.size();
        set<pair<int, int>> visit;
        priority_queue<vector<int>, vector<vector<int>>, greater<>> minHeap;
        vector<vector<int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        minHeap.push({grid[0][0], 0, 0});
        visit.insert({0, 0});

        while (!minHeap.empty()) {
            auto curr = minHeap.top();
            minHeap.pop();
            int t = curr[0], r = curr[1], c = curr[2];
            if (r == N - 1 && c == N - 1) {
                return t;
            }
            for (const auto& dir : directions) {
                int neiR = r + dir[0], neiC = c + dir[1];
                if (neiR < 0 || neiC < 0 || neiR == N || neiC == N || visit.count({neiR, neiC})) {
                    continue;
                }
                visit.insert({neiR, neiC});
                minHeap.push({max(t, grid[neiR][neiC]), neiR, neiC});
            }
        }

        return -1;
    }
};
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