On an N x Ngrid, each squaregrid[i][j]represents the elevation at that point(i,j).

Now rain starts to fall. At timet, the depth of the water everywhere ist. 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 t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square(0, 0). What is the least time until you can reach the bottom right square(N-1, N-1)?

Example 1:

Input:
 [[0,2],[1,3]]

Output:
 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.

Example 2:

Input:
 [[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]]

Output:
 16

Explanation:
 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


The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Note:

1. 2 < = N < = 50 .
2. grid[i][j] is a permutation of [0, ..., N*N - 1].

tag: priority queue

class Solution {
    public int swimInWater(int[][] grid) {
        int N = grid.length;
        Set<Integer> seen = new HashSet();
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>((k1, k2) ->
                grid[k1 / N][k1 % N] - grid[k2 / N][k2 % N]);
        pq.offer(0);
        int ans = 0;

        int[] dr = new int[]{1, -1, 0, 0};
        int[] dc = new int[]{0, 0, 1, -1};

        while (!pq.isEmpty()) {
            int k = pq.poll();
            int r = k / N, c = k % N;
            ans = Math.max(ans, grid[r][c]);
            if (r == N-1 && c == N-1) return ans;

            for (int i = 0; i < 4; ++i) {
                int cr = r + dr[i], cc = c + dc[i];
                int ck = cr * N + cc;
                if (0 <= cr && cr < N && 0 <= cc && cc < N && !seen.contains(ck)) {
                    pq.offer(ck);
                    seen.add(ck);
                }
            }
        }

        throw null;
    }
}

union-find

class Solution {
    class UnionFind {
        int[] parent;
        int[] rank;

        UnionFind(int[][] grid) {
            int N = grid.length;
            parent = new int[N * N];
            rank = new int[N * N];
            for (int i = 0; i < N * N; i++) {
                parent[i] = i;
            }
        }

        public int find(int i) {
            if (parent[i] != i) {
                // path compression
                parent[i] = find(parent[i]);
            }
            return parent[i];
        }

        public void union(int i, int j) {
            int rooti = find(i);
            int rootj = find(j);
            if (rooti == rootj) return;
            // union by rank
            if (rank[rooti] < rank[rootj]) {
                parent[rooti] = rootj;
            } else if (rank[rootj] < rank[rooti]) {
                parent[rootj] = rooti;
            } else {
                parent[rooti] = rootj;
                rank[rootj]++;
            }
        }

        public boolean isConnected() {
            return find(0) == find(parent.length - 1);
        }
    }

    private int[][] dirs = {{-1,0}, {1,0}, {0,-1}, {0,1}};

    public int swimInWater(int[][] grid) {
        UnionFind uf = new UnionFind(grid);
        int N = grid.length;
        int[][] indexes = new int[N * N][2];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                indexes[grid[i][j]][0] = i;
                indexes[grid[i][j]][1] = j;
            }
        }
        for (int i = 0; i < N * N; i++) {
            int[] p = indexes[i];
            for (int[] d : dirs) {
                int x = p[0] + d[0], y = p[1] + d[1];
                if (x < 0 || x >= N || y < 0 || y >=N || grid[x][y] > i) continue;
                // union with smaller neighbors
                uf.union(p[0] * N + p[1], x * N + y);
            }
            if (uf.isConnected()) return i;
        }
        return N * N; // dummy. will not be executed
    }
}