305. 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:

Input:
 m = 3, n = 3, positions = [[0,0], [0,1], [1,2], [2,1]]

Output:
 [1,1,2,3]

Explanation:

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

Follow up:

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

Solution

(1) Java

class Solution {
    class UF {
        int[] parent;
        int[] rank;
        public UF(int size) {
            parent = new int[size];
            rank = new int[size];
            for (int i = 0; i < size; i++) {
                parent[i] = i;
            }
        }

        public int find(int i) {
            while (i != parent[i]) {
                i = find(parent[parent[i]]);
            }
            return i;
        }

        public void union(int i, int j) {
            int pi = find(i);
            int pj = find(j);
            if (pi == pj) {
                return;
            }
            int ri = rank[pi];
            int rj = rank[pj];
            if (ri >= rj) {
                parent[pj] = pi;
                rank[pi]++;
            } else {
                parent[pi] = pj;
                rank[pj]++;
            }
        }
    }

    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        UF uf = new UF(m*n);
        List<Integer> rst = new ArrayList<>();
        int num = 0;
        int[][] grid = new int[m][n];
        int[][] dir = new int[][]{{-1,0},{1,0},{0,-1},{0,1}};
        for (int[] position : positions) {
            num++;
            int i = position[0];
            int j = position[1];
            int pos = i*n+j;
            grid[i][j] = 1;
            for (int k = 0; k < 4; k++) {
                int ni = i+dir[k][0];
                int nj = j+dir[k][1];
                if (ni >= 0 && ni < m && nj >= 0 && nj < n && grid[ni][nj] == 1) {
                    int npos = ni*n+nj;
                    int rpos = uf.find(pos);
                    int rnpos = uf.find(npos);
                    if (rpos != rnpos) {
                        uf.union(pos, npos);
                        num--;
                    }
                }
            }
            rst.add(num);
        }
        return rst;
    }
}

305. Number of Islands II