778 Swim in Rising Water

778. Swim in Rising Water

1. Question

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].

2. Implementation

(1) Union Find

class Solution {
    public int swimInWater(int[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int n = grid.length;
        int time = 0;

        UnionFind uf = new UnionFind(n * n);

        while (!uf.isConnected(0, n * n - 1)) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (grid[i][j] > time) continue;

                    if (i < n - 1 && grid[i + 1][j] <= time) {
                        uf.union(i * n + j, (i + 1) * n + j);
                    }

                    if (j < n - 1 && grid[i][j + 1] <= time) {
                        uf.union(i * n + j, i * n + j + 1);
                    }
                }
            }
            ++time;
        }
        return time - 1;
    }

    class UnionFind {
        int[] parents;
        int[] size;

        public UnionFind(int n) {
            parents = new int[n];
            size = new int[n];

            for (int i = 0; i < n; i++) {
                parents[i] = i;
                size[i] = 1;
            }
        }

        public int find(int id) {
            while (id != parents[id]) {
                id = parents[id];
            }
            return id;
        }

        public boolean isConnected(int i, int j) {
            return find(i) == find(j);
        }

        public void union(int i, int j) {
            int rootI = find(i);
            int rootJ = find(j);

            if (rootI == rootJ) return;

            if (size[rootI] < size[rootJ]) {
                parents[rootI] = rootJ;
                size[rootJ] += size[rootI];
            }
            else {
                parents[rootJ] = rootI;
                size[rootI] += size[rootJ];
            }
        }
    }
}

3. Time & Space Complexity

Union Find: 时间复杂度O(n^2 * T), T为从(0, 0) 到(n - 1, n - 1)的时间, 空间复杂度O(n^2)