305. Number of Islands II

• Difficulty: Hard

• Topics: Union Find

• Similar Questions:

  • Number of Islands

Problem:

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?

Solutions:

class Solution {
public:
    vector<int> numIslands2(int m, int n, vector<vector<int>>& positions) {
        UF uf;
        vector<int> ret;
        vector<vector<bool>> grid (m, vector<bool> (n, false));

        for (auto& position : positions) {
            int row = position[0];
            int col = position[1];
            if (grid[row][col]) {
                ret.push_back(uf.getSize());
                continue;
            }

            grid[row][col] = true;
            uf.add(row * n + col);
            if (row + 1 < m && grid[row + 1][col]) {
                uf.connect(row * n + col, (row + 1) * n + col);
            }
            if (row - 1 >= 0 && grid[row - 1][col]) {
                uf.connect(row * n + col, (row - 1) * n + col);
            }
            if (col + 1 < n && grid[row][col + 1]) {
                uf.connect(row * n + col, row * n + col + 1);
            }
            if (col - 1 >= 0 && grid[row][col - 1]) {
                uf.connect(row * n + col, row * n + col - 1);
            }

            ret.push_back(uf.getSize());
        }

        return ret;
    }

    class UF {
    public:
        int getSize() {
            return size;
        }

        void add(int a) {
            parents[a] = a;
            ++size;
        }

        void connect(int a, int b) {
            int rootA = find(a);
            int rootB = find(b);

            if (rootA != rootB) {
                parents[rootA] = rootB;
                --size;
            }
        }

        int find(int x) {
            if (parents[x] != x) {
                parents[x] = find(parents[x]);
            }
            return parents[x];
        }
    private:
        map<int, int> parents;
        int size = 0;
    };
};