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:

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?

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

public List<Integer> numIslands2(int m, int n, int[][] positions) {

List<Integer> result = new ArrayList<>();

if(m <= 0 || n <= 0) return result;

int count = 0; // number of islands

int[] roots = new int[m * n]; // one island = one tree

Arrays.fill(roots, -1);

for(int[] p : positions) {

int root = n * p[0] + p[1]; // assume new point is isolated island

roots[root] = root; // add new island

count++;

for(int[] dir : dirs) {

int x = p[0] + dir[0];

int y = p[1] + dir[1];

int nb = n * x + y;

if(x < 0 || x >= m || y < 0 || y >= n || roots[nb] == -1) continue;

int rootNb = findIsland(roots, nb);

if(root != rootNb) { // if neighbor is in another island

roots[root] = rootNb; // union two islands

root = rootNb; // current tree root = joined tree root

count--;

}

}

result.add(count);

}

return result;

}

public int findIsland(int[] roots, int id) {

while(id != roots[id]) {

//roots[id] = roots[roots[id]]; // add this line for PATH COMPRESSION

id = roots[id];

}

return id;

}