<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 getIndex(int x, int y, int n) {
        return x * n + y;
    }
    bool is_valid(int x, int y, int n) {
        return x >= 0 && x < n && y >= 0 && y < n;
    }
    void dijkstra(const vector<vector<pair<int, int>>>& graph, vector<int>& dis, int source, int init_val) {
        int n = graph.size();
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
        dis[source] = init_val;
        pq.push(make_pair(dis[source], source));
        while(!pq.empty()) {
            auto [d, u] = pq.top();
            pq.pop();
            if (d > dis[u]) continue;
            dis[u] = d;
            for(auto item : graph[u]) {
                auto [v, w] = item;
                if(max(dis[u], w) < dis[v]) {
                    dis[v] = max(dis[u], w);
                    pq.push(make_pair(dis[v], v));
                }
            }
        }
    }
    int swimInWater(vector<vector<int>>& grid) {
        const int rem_x[4] = {-1, 1, 0, 0};
        const int rem_y[4] = {0, 0, -1, 1};
        int n = grid.size();
        vector<vector<pair<int, int>>> graph(n * n, vector<pair<int, int>>());
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                int index = getIndex(i, j, n);
                for(int k = 0; k < 4; k++) {
                    int new_i = i + rem_x[k];
                    int new_j = j + rem_y[k];
                    if(!is_valid(new_i, new_j, n)) continue;
                    int new_index = getIndex(new_i, new_j, n);
                    graph[index].push_back(make_pair(new_index, grid[new_i][new_j]));
                    graph[new_index].push_back(make_pair(index, grid[i][j]));
                }
            }
        }
        vector<int> dis(n*n, INT_MAX);
        dijkstra(graph, dis, getIndex(0, 0, n), grid[0][0]);
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                int index = getIndex(i, j, n);
                cout << i << " " << j << " " << dis[index] << endl;
            }
        }
        return dis[getIndex(n - 1, n - 1, n)];
    }
};
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