Leetcode: Number of Islands II

A 2d grid map of m rows and n columns is initially filled with water. We may perform an addLand operation which turns the water at position (row, col) into a land. Given a list of positions to operate, count the number of islands after each addLand operation. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example:

Given m = 3, n = 3, positions = [[0,0], [0,1], [1,2], [2,1]].
Initially, the 2d grid grid is filled with water. (Assume 0 represents water and 1 represents land).

0 0 0
0 0 0
0 0 0
Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land.

1 0 0
0 0 0   Number of islands = 1
0 0 0
Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land.

1 1 0
0 0 0   Number of islands = 1
0 0 0
Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land.

1 1 0
0 0 1   Number of islands = 2
0 0 0
Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land.

1 1 0
0 0 1   Number of islands = 3
0 1 0
We return the result as an array: [1, 1, 2, 3]

Challenge:

Can you do it in time complexity O(k log mn), where k is the length of the positions?

 1 public class Solution {
 2     public List<Integer> numIslands2(int m, int n, int[][] positions) {
 3         int[][] dirs = new int[][]{{-1,0},{1,0},{0,1},{0,-1}};
 4         unionFind uf = new unionFind(m*n);
 5         List<Integer> res = new ArrayList<Integer>();
 6         for (int[] pos : positions) {
 7             int cur = pos[0]*n + pos[1];
 8             uf.ids[cur] = cur;
 9             uf.count++;
10             for (int[] dir : dirs) {
11                 int x = dir[0] + pos[0];
12                 int y = dir[1] + pos[1];
13                 int nb = x*n+y;
14                 if (x<0 || x>=m || y<0 || y>=n || uf.ids[nb]==-1) continue;
15                 if (uf.find(nb) != uf.find(cur)) {
16                     uf.union(nb, cur);
17                 }
18             }
19             res.add(uf.count);
20         }
21         return res;
22     }
23     
24         public class unionFind {
25             int[] ids;
26             int count;
27             public unionFind(int num) {
28                 this.ids = new int[num];
29                 Arrays.fill(ids, -1);
30                 this.count = 0;
31             }
32             public int find(int num) {
33                 return ids[num];
34             }
35             public boolean union(int n1, int n2) {
36                 int id1=ids[n1], id2=ids[n2];
37                 if (id1 != id2) {
38                     for (int i=0; i<ids.length; i++) {
39                         if (ids[i] == id2) {
40                             ids[i] = id1;
41                         }
42                     }
43                     count--;
44                     return true;
45                 }
46                 return false;
47             }
48         }
49 }

 1 int[][] dirs = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
 2 
 3 public List<Integer> numIslands2(int m, int n, int[][] positions) {
 4     List<Integer> result = new ArrayList<>();
 5     if(m <= 0 || n <= 0) return result;
 6 
 7     int count = 0;                      // number of islands
 8     int[] roots = new int[m * n];       // one island = one tree
 9     Arrays.fill(roots, -1);            
10 
11     for(int[] p : positions) {
12         int root = n * p[0] + p[1];     // assume new point is isolated island
13         roots[root] = root;             // add new island
14         count++;
15 
16         for(int[] dir : dirs) {
17             int x = p[0] + dir[0]; 
18             int y = p[1] + dir[1];
19             int nb = n * x + y;
20             if(x < 0 || x >= m || y < 0 || y >= n || roots[nb] == -1) continue;
21 
22             int rootNb = findIsland(roots, nb);
23             if(roots[root] != rootNb) {        // if neighbor is in another island
24                 roots[root] = rootNb;   // union two islands 
25                 root = rootNb;          // current tree root = joined tree root
26                 count--;               
27             }
28         }
29 
30         result.add(count);
31     }
32     return result;
33 }
34 
35 public int findIsland(int[] roots, int id) {
36     while(id != roots[id]) id = roots[id];
37     return id;
38 }